# Derive the error for the Midpoint rule

### Problem statement

Let $$[a,b]$$ be divided into $$n$$ subintervals by $$n+1$$ points $$x_0 = a, x_1 = a+h , x_2 = a+2h , \dots , x_n = a+nh =b$$.

Show that $$I- I_M = \frac{1}{24}(b-a)h^2 f''(z)$$ for some $$z \in [a,b]$$.

### Attempt at solution

$$I_M = h \sum_{i=0}^{n-1} f\left( a + \left( i + \frac12\right)\right) := h\sum_{i=0}^{n-1}f(x_i)$$

\begin{align} I -I_M &= \int_a^b f(x) \mathrm{d}x - h\sum_{i=0}^{n-1}f(x_i)\\ &= \sum_{i=0}^{n-1} \left( \int_{x_i}^{x_{i+1}} f(x) \ \mathrm{d}x - hf(x_i)\right) \end{align} Here I am stuck. I know I need to use Taylor's theorem in conjunction with Mean-Value Theorem but I don't know how to proceed.

I suspect that for $$x\in [x_i ,x_{i+1}]$$ the Taylor expansion is $$f(x) = f(x_i) + (x-x_i)f'(x_i) + \frac12 (x-x_i)^2 f''(z_x) \quad \text{for } z_x \in [x_i , x_{i+1}]$$ but then If I substitute this in the above I get \begin{align} &= \sum_{i=0}^{n-1}\left( \int_{x_i}^{x_{i+1}} (f(x_i) + (x-x_i)f'(x_i) + \frac12 (x-x_i)^2 f''(z_x))\mathrm{d}x - hf(x_i)\right)\\ \end{align} and I don't see what I have accomplished

Does anyone know If I am on the right track and if yes how to continue this derivation ?

You can start by obtaining the rule for a single interval, say \begin{align*} \int_{x_i}^{x_{i+1}} f(x) dx -hf(c_i) &= \int_{x_i}^{x_{i+1}}(f(x)-f(c_i))dx\\ & = \int_{x_i}^{x_{i+1}} \left(f(c_i)+f'(c_i)(x-c_i) + \frac{f''(\xi_i(x))}{2}(x-c_i)^2\;-\;f(c_i)\right) dx\\ & = \frac{f''(\xi_i)}{2}\int_{x_i}^{x_{i+1}}(x-c_i)^2 dx = \frac{f''(\xi_i) h^3}{24} \end{align*}

Now, if you take the sum over the $$n$$ intervals, you get the error term

$$\sum_{i=1}^n \frac{f''(\xi_i) h^3}{24} = \frac{h^2(b-a)}{24}\left(\frac 1n \sum_{i=1}^n f''(\xi_i)\right) = \frac{h^2(b-a) f''(\xi)}{24}.$$

Some notes:

1. I used the mean value theorem for integrals. Since $$(x-c_i)^2$$ does not change sign, we know that $$\int_{x_i}^{x_{i+1}} \frac{f''(\xi_i(x))}{2}(x-c_i)^2 dx = \frac{f''(\xi_i)}{2} \int_{x_i}^{x_{i+1}} (x-c_i)^2 dx$$

2. If we assume the $$f\in C^2$$, the term $$\frac 1n \sum f''(\xi_i)$$ is an average of values of $$f''$$, hence between the minimum and maximum of $$f''$$. The intermediate value theorem says that it will correspond to $$f''(\xi)$$, for some $$\xi \in (a,b)$$.

• What do you know about the function $x \mapsto \xi_i(x)$?
– RRL
Dec 8, 2021 at 12:05
• I am in fact assuming that $f''(\xi(x))$ is continuous and should have added some comment on that. Over the years I've learned that, in pedagogic terms, it is preferable to streamline the proof, leaving the technical aspects to notes or auxiliary lemmas. Dec 8, 2021 at 12:58
• This is what I was looking for thank you. Dec 8, 2021 at 18:52
• Where does the $f'(c_i) (x-c_i)$ term go in the second line ? Dec 8, 2021 at 19:43
• @hexaquark The term vanishes during the integration because it is an odd function with respect tot the midpoint $c_j$ of the interval. Dec 9, 2021 at 12:08

There is another proof that avoids double integrals and uses the generalized mean value instead. I have learned this technique from reading @LutzLehmann's answers to very similar questions.

Let $$h>0$$ be given and consider the problem of computing the integral $$I = \int_{-h}^h f(x)dx.$$ The midpoint rule takes the form $$M_h = 2h f(0).$$ We will now obtain the familiar error formula by studying the auxiliary function $$g$$ given by $$g(x) = \int_{-x}^x f(t) dt - 2x f(0).$$ This function is interesting precisely because $$g(h) = I - M_h.$$ Let $$F$$ denote an anti-derivative of $$f$$. Then $$g(x) = F(x) - F(-x) - 2xf(0).$$ It is clear that $$g$$ is as smooth as $$F$$ and that $$g(0) = 0$$. We have $$g'(x) = f(x) + f(-x) - 2f(0).$$ It follows that $$g'(0) = 0.$$ We have $$g''(x) = f'(x) - f'(-x).$$ It follows that $$g''(0) = 0.$$ We conclude that the graph of $$g$$ is quite flat near $$x=0$$. To make further progress we make repeated use of the generalized mean value theorem. There exists at least one $$x_1$$ between $$0$$ and $$x$$ such that $$\frac{g(x)}{x^3} = \frac{g(x) - g(0)}{x^3 - 0^3} = \frac{g'(x_1)}{3x_1^2}.$$ There exists at least one $$x_2$$ between $$0$$ and $$x_1$$ such that $$\frac{g'(x_1)}{3x_1^2} = \frac{g'(x_1) - g'(0)}{3x_1^2 - 3\cdot 0^2}= \frac{g''(x_2)}{6x_2}.$$ Similarly, there exists at least one $$x_3$$ between $$0$$ and $$x_2$$ such that $$\frac{g''(x_2)}{6x_2} = \frac{g''(x_2)- g''(0)}{6x_2 - 6 \cdot 0} = \frac{g'''(x_3)}{6} = \frac{f''(x_3) + f''(-x_3)}{6}$$ Finally, by the continuity of $$f''$$ there is at least one $$\xi$$ between $$-x_3$$ and $$x_3$$ such that $$\frac{g(x)}{x^3} = \frac{f''(\xi)}{3}.$$ In particular, there is at least one $$\xi$$ in $$[-h,h]$$ such that $$I - M_h = g(h) = \frac{f''(\xi)}{3} h^3.$$ Replacing $$h$$ with $$h/2$$ yields the familiar result, i.e, $$\int_{-h/2}^{h/2} f(x) dx - h f(0) = \frac{f''(\xi)}{24} h^3.$$ We finish the proof as suggested by @RRL.

To make any progress we need to consider the smoothness of $$f$$. Here I assume $$f \in C^2([a,b])$$. Denote the partition points as $$x_j = a + hj$$ and the midpoints as $$c_j = a + \frac{2j+1}{2}h$$ for $$j = 0,1,\ldots, n-1$$.

We first find the local error for a subinterval $$[x_j,x_{j+1}]$$ using the Taylor expansion with integral remainder

$$f(x) = f(c_j) +f'(c_j)(x-c_j) + \int_{c_j}^x(x-t)f''(t)\, dt$$

Integrating over $$[x_j,x_{j+1}]$$ and using $$x_{j+1} - x_j = h$$, we get

$$\tag{1}\int_{x_j}^{x_{j+1}}f(x) \, dx = hf(c_j)+ f'(c_j)\int_{x_j}^{x_{j+1}}(x-c_j) \, dx + \int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx$$

The second term on the RHS of (1) is

$$f'(c_j)\int_{x_j}^{x_{j+1}}(x-c_j) \, dx =f'(c_j) \left.\frac{(x-c_j)^2}{2} \right|_{x_j}^{x_{j+1}}= \frac{f'(c_j)}{2} \left[(x_{j+1}-c_j)^2- (x_j - c_j)^2 \right]\\ = \frac{f'(c_j)}{2}\left[\left(\frac{h}{2}\right)^2 - \left(-\frac{h}{2}\right)^2\right]= 0$$

Substituting into (1) and rearranging, we get

$$\tag{2}\int_{x_j}^{x_{j+1}}f(x) \, dx - hf(c_j)= \int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx$$

In order to apply the mean value theorem to extract the second derivative from the integral on the RHS, some manipulation with indicator functions is required to eliminate the variable integration limit. Note that

$$\int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx \\= \int_{c_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx + \int_{x_j}^{c_j} \int_{x}^{c_j}(t-x)f''(t)\, dt\, dx \\ = \int_{c_j}^{x_{j+1}} \int_{c_j}^{x_{j+1}}\mathbf{1}_{}(t \leqslant x)(x-t)f''(t)\, dt\, dx + \int_{x_j}^{c_j} \int_{x_j}^{c_j}\mathbf{1}_{}(t \geqslant x)(t-x)f''(t)\, dt\, dx$$

Now we can apply Fubini's theorem to obtain

$$\tag{3}\int_{x_j}^{x_{j+1}} \int_{c_j}^x(x-t)f''(t)\, dt\, dx \\ = \int_{c_j}^{x_{j+1}} \int_{c_j}^{x_{j+1}}\mathbf{1}_{(t \leqslant x)}(x-t)f''(t)\, dx\, dt + \int_{x_j}^{c_j} \int_{x_j}^{c_j}\mathbf{1}_{(t \geqslant x)}(t-x)f''(t)\, dx\, dt \\ = \int_{c_j}^{x_{j+1}} f''(t)\left(\int_{t}^{x_{j+1}}(x-t)\, dx\right)\, dt + \int_{x_j}^{c_j} f''(t)\left(\int_{x_j}^{t}(t-x)\, dx\right)\, dt\\ = \int_{c_j}^{x_{j+1}} f''(t)\frac{(x_{j+1}-t)^2}{2}\, dt + \int_{x_j}^{c_j} f''(t)\frac{(t-x_j)^2}{2}\, dt$$

By the mean value theorem for integrals there are points $$z_j^+ \in [c_j,x_{j+1}]$$ and $$z_j^- \in [x_j,c_j]$$ such that

$$\tag{4}\int_{c_j}^{x_{j+1}} f''(t)\frac{(x_{j+1}-t)^2}{2}\, dt + \int_{x_j}^{c_j} f''(t)\frac{(t-x_j)^2}{2}\, dt\\= f''(z_j^+)\int_{c_j}^{x_{j+1}} \frac{(x_{j+1}-t)^2}{2}\, dt + f''(z_j^-)\int_{x_j}^{c_j} \frac{(t-x_j)^2}{2}\, dt\\ = f''(z_j^+)\frac{h^3}{48} + f''(z_j^-) \frac{h^3}{48}$$

Substituting into (2) with (3) and (4) yields

$$\int_{x_j}^{x_{j+1}}f(x) \, dx - hf(c_j)=\frac{h^3}{48}\left(f''(z_j^+)+f''(z_j^-)\right)$$

Summing from $$j=0$$ to $$j = n-1$$, we get

$$I-I_M = \int_a^b f(x) \, dx - h\sum_{j=0}^{n-1}f(c_j)= \frac{h^3}{48}\sum_{j=0}^{n-1}\left(f''(z_j^+)+f''(z_j^-)\right)\\ = \frac{2nh^3}{48}\frac{1}{n}\sum_{j=0}^{n-1}\frac{f''(z_j^+)+f''(z_j^-)}{2},$$

and since $$nh = (b-a)$$,

$$\tag{5}I - I_M = \frac{(b-a)h^2}{24} \sum_{j=0}^{n-1}\frac{f''(z_j^+)+f''(z_j^-)}{2n}$$

I'll leave it to you to consider if there exists $$z \in [a,b]$$ such that the second derivative at this point equals the weighted average of second derivatives, i.e.,

$$f''(z) = \sum_{j=0}^{n-1}\frac{f''(z_j^+)+f''(z_j^-)}{2n},$$

and, hence,

$$I - I_M = \frac{(b-a)h^2}{24}f''(z)$$

Nevertheless, if the second derivative is bounded as $$|f''(x)| \leqslant \|f''||_{\infty}$$ for all $$x \in [a,b]$$, then we obtain from (5) the well-known midpoint rule error bound

$$|I - I_M| \leqslant \frac{\|f''\|_{\infty}(b-a)h^2}{24}$$

• The proof is correct but the same result can be obtained in a much simpler way just by using a second order Taylor expansion and the mean value theorem for integrals Dec 8, 2021 at 11:40
• @PierreCarre: Understood. I intentionally use the integral form of the remainder to avoid introducing another function $\xi_i(x)$ with the Lagrange form. What do you know about this function and $f’’ \circ \xi_i$. You assume it is continuous without proof as you then apply the integral MVT. So it is “simpler” because some justification is missing.
– RRL
Dec 8, 2021 at 11:58
• You are correct, leaving out the analysis of $f''\circ \xi_i$ hides the more technical details. Dec 8, 2021 at 13:00


# Documentation

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# spline

Cubic spline data interpolation

## Syntax

s = spline(x,y,xq)
pp = spline(x,y)

## Description

example

s = spline(x,y,xq) returns a vector of interpolated values s corresponding to the query points in xq. The values of s are determined by cubic spline interpolation of x and y.

example

pp = spline(x,y) returns a piecewise polynomial structure for use by ppval and the spline utility unmkpp.

## Examples

collapse all

Use spline to interpolate a sine curve over unevenly-spaced sample points.

x = [0 1 2.5 3.6 5 7 8.1 10];
y = sin(x);
xx = 0:.25:10;
yy = spline(x,y,xx);
plot(x,y,'o',xx,yy)

Use clamped or complete spline interpolation when endpoint slopes are known. This example enforces zero slopes at the end points of the interpolation.

x = -4:4;
y = [0 .15 1.12 2.36 2.36 1.46 .49 .06 0];
cs = spline(x,[0 y 0]);
xx = linspace(-4,4,101);
plot(x,y,'o',xx,ppval(cs,xx),'-');

Extrapolate a data set to predict population growth.

Create two vectors to represent the census years from 1900 to 1990 (t) and the corresponding United States population in millions of people (p).

t = 1900:10:1990;
p = [ 75.995  91.972  105.711  123.203  131.669 ...
150.697 179.323  203.212  226.505  249.633 ];

Extrapolate and predict the population in the year 2000 using a cubic spline.

spline(t,p,2000)
ans = 270.6060

Generate the plot of a circle, with the five data points y(:,2),...,y(:,6) marked with o's. The matrix y contains two more columns than does x. Therefore, spline uses y(:,1) and y(:,end) as the endslopes. The circle starts and ends at the point (1,0), so that point is plotted twice.

x = pi*[0:.5:2];
y = [0  1  0 -1  0  1  0;
1  0  1  0 -1  0  1];
pp = spline(x,y);
yy = ppval(pp, linspace(0,2*pi,101));
plot(yy(1,:),yy(2,:),'-b',y(1,2:5),y(2,2:5),'or'), axis equal

Use spline to sample a function over a finer mesh.

Generate sine and cosine curves for a few values between 0 and 1. Use spline interpolation to sample the functions over a finer mesh.

x = 0:.25:1;
Y = [sin(x); cos(x)];
xx = 0:.1:1;
YY = spline(x,Y,xx);
plot(x,Y(1,:),'o',xx,YY(1,:),'-')
hold on
plot(x,Y(2,:),'o',xx,YY(2,:),':')
hold off

Compare the interpolation results produced by spline and pchip for two different functions.

Create vectors of x values, function values at those points y, and query points xq. Compute interpolations at the query points using both spline and pchip. Plot the interpolated function values at the query points for comparison.

x = -3:3;
y = [-1 -1 -1 0 1 1 1];
xq1 = -3:.01:3;
p = pchip(x,y,xq1);
s = spline(x,y,xq1);
plot(x,y,'o',xq1,p,'-',xq1,s,'-.')
legend('Sample Points','pchip','spline','Location','SouthEast')

In this case, pchip is favorable since it does not oscillate as freely between the sample points.

Perform a second comparison using an oscillatory sample function.

x = 0:25;
y = besselj(1,x);
xq2 = 0:0.01:25;
p = pchip(x,y,xq2);
s = spline(x,y,xq2);
plot(x,y,'o',xq2,p,'-',xq2,s,'-.')
legend('Sample Points','pchip','spline')

When the underlying function is oscillatory, spline captures the movement between points better than pchip.

## Input Arguments

collapse all

x-coordinates, specified as a vector. The vector x specifies the points at which the data y is given. The elements of x must be unique.

Data Types: single | double

Function values at x-coordinates, specified as a numeric vector, matrix, or array. x and y must have the same length.

If y is a matrix or array, then the values in the last dimension, y(:,...,:,j), are taken as the values to match with x. In that case, the last dimension of y must be the same length as x.

The endslopes of the cubic spline follow these rules:

• If x and y are vectors of the same size, then the not-a-knot end conditions are used.

• If x or y is a scalar, then it is expanded to have the same length as the other and the not-a-knot end conditions are used.

• If y is a vector that contains two more values than x has entries, then spline uses the first and last values in y as the endslopes for the cubic spline. For example, if y is a vector, then:

• f(x) = y(2:end-1)

• df(min(x)) = y(1)

• df(max(x)) = y(end)

• If y is a matrix or an N-dimensional array with size(y,N) equal to length(x)+2, then these properties hold:

• f(x(j)) matches the value y(:,...,:,j+1) for j = 1:length(x)

• Df(min(x)) matches y(:,:,...:,1)

• Df(max(x)) matches y(:,:,...:,end)

Data Types: single | double

Query points, specified as a vector. The points specified in xq are the x-coordinates for the interpolated function values s that spline computes.

Data Types: single | double

## Output Arguments

collapse all

Interpolated values at query points, returned as a vector, matrix, or array.

The size of s is related to the sizes of the inputs:

• If y is a scalar or vector, then s has the same size as xq.

• If y is an array that is not a vector, then these conditions apply:

• If xq is a scalar or vector, then size(s) equals [d1, d2, ..., dk, length(xq)].

• If xq is an array of size [m1,m2,...,mj], then size(s) equals [d1,d2,...,dk,m1,m2,...,mj].

Piecewise polynomial, returned as a structure. Use this structure with the ppval function to evaluate the piecewise polynomial at one or more query points. The structure has these fields.

FieldDescription
form

'pp' for piecewise polynomial

breaks

Vector of length L+1  with strictly increasing elements that represent the start and end of each of L intervals

coefs

L-by-k  matrix with each row coefs(i,:) containing the local coefficients of an order k polynomial on the ith interval, [breaks(i),breaks(i+1)]

pieces

Number of pieces, L

order

Order of the polynomials

dim

Dimensionality of target

Since the polynomial coefficients in coefs are local coefficients for each interval, you must subtract the lower endpoint of the corresponding knot interval to use the coefficients in a conventional polynomial equation. In other words, for the coefficients [a,b,c,d] on the interval [x1,x2], the corresponding polynomial is

$f\left(x\right)=a{\left(x-{x}_{1}\right)}^{3}+b{\left(x-{x}_{1}\right)}^{2}+c\left(x-{x}_{1}\right)+d\text{\hspace{0.17em}}.$

## Tips

• You also can perform spline interpolation using the interp1 function with the command interp1(x,y,xq,'spline'). While spline performs interpolation on rows of an input matrix, interp1 performs interpolation on columns of an input matrix.

## Algorithms

A tridiagonal linear system (possibly with several right-hand sides) is solved for the information needed to describe the coefficients of the various cubic polynomials that make up the interpolating spline. spline uses the functions ppval, mkpp, and unmkpp. These routines form a small suite of functions for working with piecewise polynomials. For access to more advanced features, see interp1 or the Curve Fitting Toolbox™ spline functions.

## References

[1] de Boor, Carl. A Practical Guide to Splines. Springer-Verlag, New York: 1978.

## See Also

#### Introduced before R2006a

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# Proving set identities: empty set case.

I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $$A,B,C$$ we have $$A\cap(B \cup C) = (A \cap B) \cup (A \cap C).$$

I understand that this can be done by proving both $$A\cap(B \cup C) \subset (A \cap B) \cup (A \cap C) \ \text{and} \ (A \cap B) \cup (A \cap C) \subset A\cap(B \cup C)$$ hold true.

We can do this by letting $$x$$ be an arbitrary element of $$A\cap(B \cup C)$$ and showing that it is an element of $$(A \cap B) \cup (A \cap C)$$, and vice versa.

But what about when $$A \cap (B \cup C)$$ is the empty set? Then I would think we can't let $$x$$ be an arbitrary element of $$A \cap (B \cup C)$$ since there are none. However, I am aware that $$A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$$ is trivially true in this case.

In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?

If so, some suggestions as to how to incorporate them into proofs would be helpful :-).

• Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that, – DonAntonio Dec 22 '18 at 18:35
• Words to live by: "vacuous truth". I am almost tempted to say this is a duplicate of many other questions that were asked before. For example, math.stackexchange.com/questions/734418/… math.stackexchange.com/questions/2723860/… math.stackexchange.com/questions/1953218/… and there are many others which are similar. – Asaf Karagila Dec 23 '18 at 10:24

In order to show that for two sets $$X$$ and $$Y$$ it holds that $$X\subseteq Y$$, you have to prove that

for every $$x$$, if $$x\in X$$, then $$x\in Y$$.

Note that a statement of the form “if $$\mathscr{A}$$ then $$\mathscr{B}$$” is true when

either $$\mathscr{A}$$ is false or both $$\mathscr{A}$$ and $$\mathscr{B}$$ are true

If $$X$$ is the empty set, then “$$x\in X$$” is false for every $$x$$; hence “if $$x\in X$$ then $$x\in Y$$” is true.

The phrase “take an arbitrary element $$x\in X$$” is possibly misleading, but its intended meaning is “suppose $$x\in X$$”.

• Ah, so if we write "suppose $x \in X$" and successfully deduce that "$x \in Y$" , we have shown that "if $x \in X$, then $x \in Y$" is true. But, if "$x \in X$" is always false, we have still shown "if $x \in X$ then $x \in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct? – user445909 Dec 22 '18 at 19:39
• @E-mu Basically so – egreg Dec 22 '18 at 20:27

If $$D$$ is the empty set then the following statement is always true:$$\text{If }x\in D\text{ then }\cdots$$no matter what the dots are standing for.

It is false that $$x\in D$$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.

So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.

Since the empty set is a subset of any set, there is no need of including that in a formal proof. However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .


# Determine the number of subsets

How many distinct subsets of a set $\text{A}$ are there, containing at least $9$ elements, where the total number of elements in set $\text{A}$ is $18$ ?

I've solved it by making cases of either choosing $9$, $10$, $11$ and so on elements from the set and then adding them up, i.e, $$\sum_{r=9}^{18} \dbinom{18}{r}$$

However, the answer given in my book is $$\dfrac{1}{2} \times \dfrac{18!}{9! \cdot 9!} + 2^{17}$$

Although the numerical value of my answer and book's answer is equal, I'm interested in knowing the intended solution of the author.

The term $\dfrac{1}{2} \times \dfrac{18!}{9! \cdot 9!}$ is dividing $18$ elements into two equal parts. The term $2^{17}$ is the total number of subsets of a set containing $17$ elements. But I can't connect these two things with the question.

Any help will be appreciated.
Thanks.

P.S. If anyone is interested in knowing how I found $\displaystyle \sum_{r=9}^{18} \dbinom{18}{r}$ here is my method,

Let $\text{S} = \displaystyle \sum_{r=9}^{18} \dbinom{18}{r}$

By Binomial Theorem,

$2^{18} = (1+1)^{18} = \displaystyle \sum_{r=0}^{18} \dbinom{18}{r}$

$\implies 2\text{S} - \dbinom{18}{9} =2^{18} \left( \because \dbinom{18}{r} = \dbinom{18}{18-r}\right)$

and the rest follows.

If $A$ has $18$ elements, and $B\subseteq A$ has fewer than $9$ elements, then $B^c\subseteq A$ has more than nine elements. Also, if $B$ has more than $9$ elements then $B^c$ has fewer than $9$ elements. That means the number of subsets of $A$ with fewer than $9$ elements equals the number of subsets of $A$ with more than $9$ elements.
Of course, the complement of a set with $9$ elements is another set with $9$ elements. The number you want, the number of subsets of $A$ with at least $9$ elements, is almost half the number of all subsets of $A$. We can figure out just how close that almost is.
Let $m$ be the number of subsets of $A$ with more than $9$ elements, so $m$ is also the number of subsets of $A$ with fewer than $9$ elements. Let $n$ be the number of subsets of $A$ with exactly $9$ elements. Then you want to find $m+n$, and we can find it from the total number of subsets of $A$, which we'll call $t$.
$$\text{(fewer than 9)+(exactly 9)+(more than 9)=(all)}$$ $$m+n+m=t$$ \begin{align} m+n&=\frac 12n+\frac 12t \\ &=\frac 12{18 \choose 9}+\frac 12\cdot 2^{18} \\ &=\frac 12 \cdot \frac{18!}{9! \cdot 9!} + 2^{17} \end{align}
Well, there are $2^{18}$ subsets all in all. Of these, $\dfrac{18!}{9!\cdot 9!}$ have exactly $9$ elements. The rest either have more than $9$ or fewer than $9$. Now, by symmetry, the number of subsets with more than $9$ elements equals the number of subsets with fewer than $9$ (pair each subset with its complement). Thus the number of subsets with more than $9$ elements is $\dfrac{1}{2} \times \left(2^{18} - \dfrac{18!}{9!\cdot 9!}\right)$. Adding the number of subsets with exactly $9$ elements gives the desired result.


A circle with diameter the minor base $$CD$$ of a trapezium $$ABCD$$ intersects its diagonals $$AC$$ and $$BD$$ in, respectively, their midpoints $$M$$ and $$N$$. The lines $$DM$$ and $$CN$$ intersect in $$P$$ and $$AC$$ and $$BD$$ intersect in $$H$$. Show that $$AD=CD=BC$$ and $$HP \perp AB$$.

$$DMNC$$ is a cyclic quadrilateral and $$CD||MN$$, thus $$DMNC$$ is an isosceles trapezoid. Therefore, $$CM=DN$$ and $$AC=BD$$. Now I am trying to show $$AD=CD$$. $$\angle DCM$$ is inscribed and it's equal to $$\angle DNM$$ but I don't see how to compare it with $$\angle CAD$$. How is this done? For the second part of the problem, I tried to show that $$HO$$ passes through $$P$$ ($$HO \perp CD$$ because $$\triangle CDH$$ is isosceles and if we show $$P \in HO$$ we are done).

Since $$DM\perp AC$$ and $$M$$ is a midpoint of $$AC$$, we obtain $$AD=DC$$.
Since $$DMNC$$ is an isosceles trapezoid, we obtain: $$\measuredangle PMN=\measuredangle PDC=\measuredangle PCD=\measuredangle PNM,$$ which gives $$PM=PN.$$ Also, $$\Delta MDN\cong\Delta NCM,$$ which gives $$\measuredangle HMN=\measuredangle HNM,$$ which gives $$HM=HN,$$ which says that $$PMHN$$ is a kite, which says $$PH\perp MN.$$ About a kite see here: https://en.wikipedia.org/wiki/Kite_(geometry)
• Yes. Thank you! $DM$ is the perpendicular bisector of $AC$. – Andrew Rogers Aug 19 '19 at 16:34
• Now I am trying to show that $HP \perp AB$ $(H = AC \cap BD)$. Is this true for all cases, because I can't prove it? – Andrew Rogers Aug 19 '19 at 16:52
• @Andrew Rogers Yes it's true because $HM=HN$ and $PM=PN.$ It's true if $AD=BC,$ which we got. – Michael Rozenberg Aug 19 '19 at 16:54
• I am not sure I understand why it's true iff $HM=HN$ and $PM=PN$. – Andrew Rogers Aug 19 '19 at 17:01


What's the probability that a sequence of coin flips never has twice as many heads as tails?

I gave my friend this problem as a brainteaser; while her attempted solution didn't work, it raised an interesting question.

I flip a fair coin repeatedly and record the results. I stop as soon as the number of heads is equal to twice the number of tails (for example, I will stop after seeing HHT or THTHHH or TTTHHHHHH). What's the probability that I never stop?

I've tried to just compute the answer directly, but the terms got ugly pretty quickly. I'm hoping for a hint towards a slick solution, but I will keep trying to brute force an answer in the meantime.

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Hopefully whatever method resolves this applies to other ratios than $2$. –  anon Aug 27 '11 at 3:32
@anon, good idea... See my answer. –  Did Aug 27 '11 at 13:53
@anon: I've generalized my answer up to an infinite series expression as well. –  Mike Spivey Aug 27 '11 at 21:00
See "a coin toss question" for the similar problem of flipping until achieving the same number of heads as tails. –  Mike Spivey Aug 31 '11 at 20:55
@Elliot I'd like you remind you that this question doesn't currently have an accepted answer. –  Git Gud Jun 16 '13 at 9:36

The game stops with probability $u=\frac34(3-\sqrt5)=0.572949017$.

See the end of the post for generalizations of this result, first to every asymmetric heads-or-tails games (Edit 1), and then to every integer ratio (Edit 2).

To prove this, consider the random walk which goes two steps to the right each time a tail occurs and one step to the left each time a head occurs. Then the number of tails is the double of the number of heads each time the walk is back at its starting point (and only then). In other words, the probability that the game never stops is $1-u$ where $u=P_0(\text{hits}\ 0)$ for the random walk with equiprobable steps $+2$ and $-1$.

The classical one-step analysis of hitting times for Markov chains yields $2u=v_1+w_2$ where, for every positive $k$, $v_k=P_{-k}(\text{hits}\ 0)$ and $w_k=P_{k}(\text{hits}\ 0)$. We first evaluate $w_2$ then $v_1$.

The $(w_k)$ part is easy: the only steps to the left are $-1$ steps hence to hit $0$ starting from $k\ge1$, the walk must first hit $k-1$ starting from $k$, then hit $k-2$ starting from $k-1$, and so on. These $k$ events are equiprobable hence $w_k=(w_1)^k$. Another one-step analysis, this time for the walk starting from $1$, yields $$2w_1=1+w_3=1+(w_1)^3$$ hence $w_k=w^k$ where $w$ solves $w^3-2w+1=0$. Since $w\ne1$, $w^2+w=1$ and since $w<1$, $w=\frac12(\sqrt5-1)$.

Let us consider the $(v_k)$ part. The random walk has a drift to the right hence its position converges to $+\infty$ almost surely. Let $k+R$ denote the first position visited on the right of the starting point $k$. Then $R\in\{1,2\}$ almost surely, the distribution of $R$ does not depend on $k$ because the dynamics is invariant by translations, and $$v_1=r+(1-r)w_1\quad\text{where}\ r=P_{-1}(R=1).$$ Now, starting from $0$, $R=1$ implies that the first step is $-1$ hence $2r=P_{-1}(A)$ with $A=[\text{hits}\ 1 \text{before}\ 2]$. Consider $R'$ for the random walk starting at $-1$. If $R'=2$, $A$ occurs. If $R'=1$, the walk is back at position $0$ hence $A$ occurs with probability $r$. In other words, $2r=(1-r)+r^2$, that is, $r^2-3r+1=0$. Since $r<1$, $r=\frac12(3-\sqrt5)$ (hence $r=1-w$).

Plugging these values of $w$ and $r$ into $v_1$ and $w_2$ yields the value of $u$.

Edit 1 Every asymmetric random walk which performs elementary steps $+2$ with probability $p$ and $-1$ with probability $1-p$ is transient to $+\infty$ as long as $p>\frac13$ (and naturally, for every $p\le\frac13$ the walk hits $0$ with full probability). In this regime, one can compute the probability $u(p)$ to hit $0$. The result is the following.

For every $p$ in $(0,\frac13)$, $u(p)=\frac32\left(2-p-\sqrt{p(4-3p)}\right).$

Note that $u(p)\to1$ when $p\to\frac13$ and $u(p)\to0$ when $p\to1$, as was to be expected.

Edit 2 Coming back to symmetric heads-or-tails games, note that, for any fixed integer $N\ge2$, the same techniques apply to compute the probability $u_N$ to reach $N$ times more tails than heads.

One gets $2u_N=E(w^{R_N-1})+w^N$ where $w$ is the unique solution in $(0,1)$ of the polynomial equation $2w=1+w^{1+N}$, and the random variable $R_N$ is almost surely in $\{1,2,\ldots,N\}$. The distribution of $R_N$ is characterized by its generating function, which solves $$(1-(2-r_N)s)E(s^{R_N})=r_Ns-s^{N+1}\quad\text{with}\quad r_N=P(R_N=1).$$ This is equivalent to a system of $N$ equations with unknowns the probabilities $P(R_N=k)$ for $k$ in $\{1,2,\ldots,N\}$. One can deduce from this system that $r_N$ is the unique root $r<1$ of the polynomial $(2-r)^Nr=1$. One can then note that $r_N=w^N$ and that $E(w^{R_N})=\dfrac{Nr_N}{2-r_N}$ hence some further simplifications yield finally the following general result.

For every $N\ge2$, $u_N=\frac12(N+1)r_N$ where $r_N<1$ solves the equation $(2-r)^Nr=1$.

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+1: Nice work, Didier! (And for the side effect of showing me that the answer I got is correct, too.) –  Mike Spivey Aug 27 '11 at 14:56
@Mike, thanks. To tell the truth, I DID compute the numerical value of my answer as soon as I got a formula, to check whether it coincided with the numerical value of your mysterious formula and I WAS relieved to see these coincided... –  Did Aug 27 '11 at 16:28
+1, I had a feeling a Markov approach would pop up. Nice to see such a simple polynomial's root solve the general form. –  anon Aug 27 '11 at 21:24
@anon, yes. This is no accident that the computations go through and that the algebra is nice, there is an underlying structure to all this, sometimes called cycles-and-weights Markov chains, which is a nifty extension of +/-1 random walks (I might add some references one day). –  Did Aug 27 '11 at 21:51

(Update: The answer to the original question is that the probability of stopping is $\frac{3}{4} \left(3 - \sqrt{5}\right)$. See end of post for an infinite series expression in the general case.)

Let $S(n)$ denote the number of ways to stop after seeing $n$ tails. Seeing $n$ tails means seeing $2n$ heads, so this would be stopping after $3n$ flips. Since there are $2^{3n}$ possible sequences in $3n$ flips, the probability of stopping is $\sum_{n=1}^{\infty} S(n)/8^n$.

To determine $S(n)$, we see that there are $\binom{3n}{n}$ ways to choose which $n$ of $3n$ flips will be tails. However, this overcounts for $n > 1$, as we could have seen twice as many heads as tails for some $k < n$. Of these $\binom{3n}{n}$ sequences, there are $S(k) \binom{3n-3k}{n-k}$ sequences of $3n$ flips in which there are $k$ tails the first time we would see twice as many heads as tails, as any of the $S(k)$ sequences of $3k$ flips could be completed by choosing $n-k$ of the remaining $3n-3k$ flips to be tails. Thus $S(n)$ satisfies the recurrence $S(n) = \binom{3n}{n} - \sum_{k=1}^{n-1} \binom{3n-3k}{n-k}S(k)$, with $S(1) = 3$.

The solution to this recurrence is $S(n) = \frac{2}{3n-1} \binom{3n}{n}.$ This can be verified easily, as substituting this expression into the recurrence yields a slight variation on Identity 5.62 in Concrete Mathematics (p. 202, 2nd ed.), namely, $$\sum_k \binom{tk+r}{k} \binom{tn-tk+s}{n-k} \frac{r}{tk+r} = \binom{tn+r+s}{n},$$ with $t = 3$, $r = -1$, $s=0$.

So the probability of stopping is $$\sum_{n=1}^{\infty} \binom{3n}{n} \frac{2}{3n-1} \frac{1}{8^n}.$$

Mathematica gives the closed form for this probability of stopping to be $$2 \left(1 - \cos\left(\frac{2}{3} \arcsin \frac{3 \sqrt{3/2}}{4}\right) \right) \approx 0.572949.$$

Added: The sum is hypergeometric and has a simpler representation. See Sasha's comments for why the sum yields this closed form solution and also why the answer is $$\frac{3}{4} \left(3 - \sqrt{5}\right) \approx 0.572949.$$

Added 2: This answer is generalizable to other ratios $r$ up to the infinite series expression. For the general $r \geq 2$ case, the argument above is easily adapted to produce the recurrence $S(n) = \binom{(r+1)n}{n} - \sum_{k=1}^{n-1} \binom{(r+1)n-(r+1)k}{n-k}S(k)$, with $S(1) = r+1$. The solution to the recurrence is $S(n) = \frac{r}{(r+1) n - 1} \binom{(r+1) n}{n}$ and can be verified easily by using the binomial convolution formula given above. Thus, for the ratio $r$, the probability of stopping has the infinite series expression $$\sum_{n=1}^{\infty} \binom{(r+1)n}{n} \frac{r}{(r+1)n-1} \frac{1}{2^{(r+1)n}}.$$ This can be expressed as a hypergeometric function, but I am not sure how to simplify it any further for general $r$ (and neither does Mathematica). It can also be expressed using the generalized binomial series discussed in Concrete Mathematics (p. 200, 2nd ed.), but I don't see how to simplify it further in that direction, either.

Added 3: In case anyone is interested, I found a combinatorial proof of the formula for $S(n)$. It works in the general $r$ case, too.

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It's late here, and I'm going to bed. Any comments, corrections, etc., will be dealt with in the morning. And if someone has an explanation for Mathematica's output I would love to see it! –  Mike Spivey Aug 27 '11 at 6:50
Wow, incredible (both the write up and the result itself)! However, I have a slight point of confusion--why is S(1) = 1, rather than 3? –  Elliott Aug 27 '11 at 7:48
Does anyone know (or perhaps have a useful reference book to find out from) the generating function of $a_n = \binom{3n}{n}$? That should help in finding the closed form of this series. –  Ragib Zaman Aug 27 '11 at 9:17
Let $u$ be you answer, then $\cos(\frac{2}{3}\arcsin \frac{3}{4}\sqrt{\frac{3}{2}}) = 1-\frac{u}{2}$. Using $\cos(3 x) = -3 \cos(x) + 4 \cos(x)^3$, and $\cos( 2 \arcsin \frac{3}{4}\sqrt{\frac{3}{2}}) = -\frac{11}{16}$, it follows $u$ satisfy a cubic with rots $u_1=\frac{3}{2}$, $u_2 = \frac{3}{4}( 3- \sqrt{5})$ and $u_3 = \frac{3}{4}(3+ \sqrt{5})$. Root $u_2$ is the one corresponding to the original expression, by numerical comparison. –  Sasha Aug 27 '11 at 13:47
@Ragib $\sum_{k\ge 0} \binom{3 k}{k} z^k$ is a hypergeometric sum with answer $\frac{3}{4 \pi} F(\frac{1}{3}, \frac{2}{3} ; \frac{1}{2} ; \frac{9 z}{4} )$. This Gauss hypergeometric function admits simpler representation. –  Sasha Aug 27 '11 at 13:57

Here's another solution, following Ross Millikan's hint:

Let $n=H−2T$ (H=heads, T=tails). At each step, $n := n+1$ or $n := n-2$ with probability 1/2. The game starts with $n=0$ and stops when $n=0$.

For any iteration (after the initial), let $P(n)$ be the probability that the game eventually stops, given the current value of $n$ (and given that the event has not yet occurred). It's clear that $P(n)$ does not depend on time. Then the following recurrence holds:

$$P(n)= \frac{P(n+1)+P(n-2)}{2} \;, \;\; n\ne 0$$

with $P(0)=1$. We also know (do we?) that $P(-\infty)=0$ (but we don't yet know $P(+\infty)$)

The solution to this difference equation (I'll spare you the details, just the usual linear difference equation procedure, in the two regions, with the above boundary conditions) is given by:

$$P(n)= \left\{ \begin{array}{ll} \phi^{-n} & \mbox{if } n \le 0 \\ 1 + B \, [(-\phi)^n -1] & \mbox{if } n \ge 0 \end{array} \right.$$

with $B= \phi^2 (1-\phi)$, $\phi = (\sqrt{5}-1)/2$

Now, after the first step we have $n=1$ or $n=-2$ with equal probability, then the probability that the game eventually stops is

$$\frac{P(1)+P(-2)}{2}=\frac{2 \phi^2 -\phi +1}{2} = 0.572949$$

The form of $P(n)$ is interesting:

This, for example, shows that the probability of stop is strongly dependent on the first coin toss (less than 40% if tail, more than 75% if head).

This procedure also seems directly generalizable, either to asymmetric coins or other integer ratios.

Added: Here goes the details for solving the recursion:

We postulate a solution of the form $P(n)= r^n$ and replacing in the recursion $2 P(n) - P(n+1) - P(n-2) = 0$ we get $$2 \; r^2 - r^3 - 1 = 0$$ The root $r_1=1$ comes immediately, and then the others: $r_2 = - \phi$, $r_3 = 1/\phi$. Then the general solution is given by $A + B (-\phi)^n + C \phi^{-n}$, for some $A,B,C$, in each zone.

In the zone $n \le 0$: we have $P(0)=1$ and $P(-\infty)=0$. As $\phi \approx 0.618 < 1$, this implies $A=0$, $B=0$,$C=1$ Hence

$$P(n) = \phi^{-n} \hspace{10px} n \le 0$$

In the zone $n \ge 0$: we have $P(0)=1$, but we don't know $P(\infty)$. We do know it's bounded, so $C=0$, and $A=1-B$, so

$$P(n) = 1 + B \, [(-\phi)^n -1] \hspace{10px} n \ge 0$$

To get rid of the remaining degree of freedom, we write the recursion for $n=1$, and replace the value of $P(-1)$ for the previous solution:

$$2 \, P(1) = P(2) + P(-1)$$ $$2 \, (1 + B \, [-\phi -1]) = 1 + B \, [(-\phi)^2 -1] + \phi$$

From this, we obtain $B= \phi^2 (1-\phi)$.

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This solution is beautiful! One thing I don't understand is, how did you set boundary conditions for the region n >= 0? –  Elliott Aug 29 '11 at 16:34
@Elliot: You first solve for $n \le 0$, then plugging it in the recursion for $n=1$ you resolve the remaining degree of freedom. Tell me if this is not clear, and I'll write the details. –  leonbloy Aug 29 '11 at 17:44
So... in the end, you know that P(-infty)=0? –  Did Aug 31 '11 at 21:01
Well, we "knew" (actually assumed) that from the beginning, it was $P(+\infty)$ the unknown. It's true (are you pointing to this?) that $P(-\infty)=0$ is intuitive and true but a rigorous proof is lacking here. –  leonbloy Aug 31 '11 at 21:19
Just pointing to the fact that, when P(-infty) first appear, you feel necessary to signal that the relation P(-infty)=0 might need a proof. But when you need the value of P(-infty) later on, you state it without further ado. Hence, reading your post, I was wondering if I had missed, somewhere inbetween these two appearances, a proof that P(-infty)=0. (Please use @.) –  Did Sep 1 '11 at 13:22

I would suggest $1-\phi=1-\frac{\sqrt{5}-1}{2}$. If define $P(n)$ as the probability that if we have an excess of tails of $n$ we will eventually have twice the heads as tails, the recurrence is $$P(n)=\frac{P(n+1)+P(n-2)}{2}$$ with boundary conditions $$P(n)=\begin {cases}1 &n\gt 0 \\ 0 &n=-\infty \end {cases}$$ The recurrence has solution $P(n)=\phi ^n$

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Sorry, I'm not sure I follow. Can you please explain what P(n) represents in slightly more detail? –  Elliott Aug 27 '11 at 6:18
+1 Nice answer, though cryptic (and slightly incorrect). To clarify: Let $n(t)=H-2T$ (H=heads, T=tails), say we are interest in the event $n(t)>0$ for some $t$. Let $P(n)$ (for $n\le0$) be the probability of that event, given the value $n$ that at that time, and that the event has not ocurred before; further, let $P(1)=1$. Then, the recursion is correct, and so is the solution (actually I think its $p(n)=\phi^{1-n}$, but anyway $P(0)= 0.618033989$). But... (continued) –  leonbloy Aug 27 '11 at 15:17
The problem is that the original game stops when $H=2T$ exactly, and this event is not equivalent to $H>2T$ (for example, the sequence $HHHTHHHTHHT...$ does not make the game stop. I wonder if this can be corrected, I liked this approach. –  leonbloy Aug 27 '11 at 15:22
@leonbloy: Good elaboration. Thanks. I was thinking that if we have excess heads then with probability 1 we get back to a 2:1 ratio. But you are right we can skip over it. A more telling (to me) sequence that fails is $HHHTTTHTHTHTHT...$ –  Ross Millikan Aug 27 '11 at 15:28
The approach was indeed fixable, I posted the derivation in my own answer - I hope this it's ok. –  leonbloy Aug 29 '11 at 0:15


# What quantifies as a rigorous proof?

Okay I have been thinking about this common combinatorial identity. $$\sum_{r=0}^{n} \binom{n}{r} = 2^n.$$ It is simple to prove this by induction, but it requires some annoying algebraic manipulation which involves other combinatorial identities. What about if we took an approach that takes into factor a specific rephrase of the problem?

My approach is this. Consider a combination lock with n number of on-off switches. To open it require turning on the correct switches. How many combinations are there? clearly, the first switch can be on or off, second on and off etc. so the combintions are $2^n$

We look at it from a different angle. How many ways are there to pick one on switch. $\dbinom{n}{1}$. what about picking 2 switches? $\dbinom{n}{2}$... and so on. Since the number of combinations are supposed to be the same,it gives us the identity.

Now i know this explanation is not rigorous, but can someone point out exactly to me exactly which part of it it fails to be so?

I have also encountered this problem in man other areas of mathematics when I try to prove something, I used a specific problem to do so and thus I doubt the validity of the proof. I can give more examples if needed

• It's a combinatorial proof. Rigorous enough for me, but I know teachers who don't like proofs without lots of algebra. – Jean-Claude Arbaut Mar 15 '13 at 14:05
• Very clever proof. To me it is the right proof. Rigor is an ambiguous concept, but it's certainly not the case that a rigorous proof of an identity needs be done algebraically. Your argument explains why the identity is true. The algebra does not give any insight. – Cheerful Parsnip Mar 15 '13 at 14:09
• I have edited your question, employing the $\LaTeX$ notation that is used on MSE. – Andreas Caranti Mar 15 '13 at 14:14
• Better use the standard $\binom{n}{k}$ notation... – vonbrand Mar 15 '13 at 14:14
• It’s a variant of the standard combinatorial proof: $2^n$ is the number of subsets of $[n]=\{1,\dots,n\}$, and $\binom{n}k$ is the number of subsets of $[n]$ of size $k$, so $2^n=\sum_k\binom{n}k$. As such it is, as others have said, just fine. And while one ought to be able to carry out the details of the proof by induction, combinatorial arguments like this one generally give much more insight into why the result is true. – Brian M. Scott Mar 15 '13 at 14:46

Your approach is an often-used (and perfectly valid) proof technique. You should be aware that some instructors don't like it, but it certainly does the job.

• An instructor who doesn’t accept combinatorial arguments ought not to be teaching the subject. – Brian M. Scott Mar 15 '13 at 14:44
• @Brian M. Scott: surely it depends on whether the goal is to teach the identity, or whether it is to teach students how to write proofs by induction. I would usually take a car if I need to travel 10 miles, but not if the point is to train for a marathon. – Carl Mummert Mar 15 '13 at 15:09
• @Carl: If the goal is to teach students how to write proofs by induction, the instructor should damned well come up with problems for which induction is the best approach. And if even then a student comes up with a different approach, the instructor should be pleased. – Brian M. Scott Mar 15 '13 at 15:11
• @Carl: My point is that the directions should not begin so. I consider it a failure on the part of the instructor if he or she has to specify a technique in order to give students practice in applying it. – Brian M. Scott Mar 15 '13 at 15:26
• I strongly agree with Brian here. Questions that prescribe techniques are either lazy pedagogy or bad curriculum design. If technique $Z$—say, induction—is a useful technique, then it is because there is some problem $Y$ such that $Z$ is superior to all other techniques for solving $Y$. If all such $Y$ are outside the scope of the class, then $Z$ is outside the scope of the class too. If, on the other hand, there is some $Y$ that is in the scope of the class, it is the instructor's job to find it and present it to the students, as an instructive example. – MJD Mar 15 '13 at 15:31

You can make such an argument rigorous. When things start to get more sophisticated, it often helps immensely to give more detail than one usually sees in this style of proof.

Let $S$ be a set with $n$ elements. Let $P(S)$ be its powerset -- that is, the set of all subsets of $S$. Let $P_i(S)$ be the set of all subsets of $S$ with exactly $i$ elements.

Your proof is then the invocation of four facts:

• The well-known formula for the number of elements in $P(S)$
• The well-known formula for the number of elements in $P_i(S)$
• The observation that $P(S)$ is the disjoint union of all of the $P_i(S)$
• The relationship between the number of elements in a sequence of sets and the number of elements in their disjoint union


My Math Forum What is the probability of getting the same number twice?
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 Probability and Statistics Basic Probability and Statistics Math Forum

 June 16th, 2017, 11:34 AM #11 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित i would think it this way: Since I have 24 draws, I would take 23 different numbers in my first 23 draws and 1 among those 23 numbers in my 24th draw. In 1st draw we can take any number but in the following draws we cannot take numbers which are already drawn, thus the numerator of the probabilities are decreasing by 1 in each of the following draw. Thus the probability of getting 23 different numbers in the first 23 draw is: $\displaystyle P1 = \frac{255}{255}\times \frac{254}{255}\times \frac{253}{255}\times ......$ till 23 steps $\displaystyle P1 = \frac{255!}{255^{23}\times 232!}$ Then for the 24th draw, we have to draw one of the numbers which we have already drawn. There are 23 possible numbers which we can draw out of 255 numbers. Thus the probability of having such number in 24th draw is: $\displaystyle P2=\frac{23}{255}$ Hence, the total probability is: $\displaystyle P=P1 \times P2$ $\displaystyle P= \frac{255!\times 23}{255^{24}\times 232!}$ $\displaystyle P\approx 0.0324$ Thanks from JeffM1
 June 16th, 2017, 11:57 AM #12 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics I'd like to present an alternative approach to the problem that could possibly shed light on that number. Consider the Markov chain with states 0, 1, ... 7, where state 0 is the initial state where no numbers have been chosen, states i, $\displaystyle 1 \leq i \leq 6$ denote the state where i distinct numbers have been chosen and no repeats have appeared so far, and 7 denote the state where repeats have appeared. Obviously, the transition matrix is as follows: $\displaystyle \mathbf{P} = \begin{bmatrix} 0 &1 &0 & 0 & 0 & 0 & 0 & 0\\ 0 &0 & \frac{5}{6} & 0 & 0& 0& 0& \frac{1}{6}\\ 0 & 0 & 0 & \frac{2}{3} & 0 & 0 &0 & \frac{1}{3}\\ 0 & 0 &0 &0 & \frac{1}{2}& 0 & 0& \frac{1}{2}\\ 0 & 0 & 0& 0 & 0 &\frac{1}{3} & 0 & \frac{2}{3}\\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{6} & \frac{5}{6}\\ 0& 0 & 0 & 0 & 0 & 0 &0 &1 \\ 0&0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ and the initial distribution is $\displaystyle \mathbf{e}_1$. So all we have to do is to calculate $\displaystyle \mathbf{e}_1\mathbf{P}^3 = (0,0,0,\frac{5}{9},0,0,0,\frac{4}{9})^T$, verifying Jeff's answer. Using a well-known theorem on absorbing states and some concepts in elementary probability (in particular the probability generating function), we can even do other cool things like determining how many numbers we need to draw on average to get a repeat, but I digress. So we can immediately apply this to the initial question: $\displaystyle \mathbf{P} =\begin{bmatrix} 0 & 1 & 0 & ... & 0 & 0\\ 0 & 0 & 254/255 & ... & 0& 1/255\\ 0 & 0 & 0 & \ddots & 0 & \vdots\\ 0& 0 & 0 & ... & 0 & 1\\ 0 & 0 & 0 & ...& 0 &1 \end{bmatrix}$ MATLAB tells us that 25th and last entries of $\displaystyle \mathbf{e}_1 \mathbf{P}^{24}$ are 0.32719 and 0.67281 respectively, again verifying Jeff's claims. Thanks from JeffM1 Last edited by 123qwerty; June 16th, 2017 at 11:59 AM.
 June 16th, 2017, 01:00 PM #13 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 I stick to my guns: the probability is ~.39
June 16th, 2017, 01:24 PM   #14
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Quote:
 Originally Posted by Denis Well, whatever... IF, IF, IF the 24 draws produce: 1 pair no other number matching that pair 22 remaining are ALL different then probability = ~.39
I agree that that is the apparent question originally asked.

How many ways can I draw 24 cards with replacement from a set of 255 cards consecutively numbered 1 through 255?

Answer: $255^{24}.$

Do you agree?

How many ways can I draw 23 differently numbered cards from that same set with replacement?

Answer: $\displaystyle \prod_{j=1}^{23}(256 - j).$

Do you agree?

How many ways can I draw one of 23 specific numbers from that set?

Answer: $23.$

Do you agree?

So the probability of drawing 23 different cards and 1 duplicate card in 24 draws with replacement should be

$\dfrac{23 * \displaystyle \prod_{j=1}^{23}(256 - j)}{255^{24}} \approx 0.0324.$

I can't get to your result.

June 16th, 2017, 03:09 PM   #15
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Quote:
 Originally Posted by JeffM1 I agree that that is the apparent question originally asked. How many ways can I draw 24 cards with replacement from a set of 255 cards consecutively numbered 1 through 255? Answer: $255^{24}.$ Do you agree? How many ways can I draw 23 differently numbered cards from that same set with replacement? Answer: $\displaystyle \prod_{j=1}^{23}(256 - j).$ Do you agree? How many ways can I draw one of 23 specific numbers from that set? Answer: $23.$ Do you agree? So the probability of drawing 23 different cards and 1 duplicate card in 24 draws with replacement should be $\dfrac{23 * \displaystyle \prod_{j=1}^{23}(256 - j)}{255^{24}} \approx 0.0324.$ I can't get to your result.
i absolutely agree with you.

 June 16th, 2017, 04:04 PM #16 Senior Member   Joined: May 2016 From: USA Posts: 825 Thanks: 335 I am impressed by 123qwerty's translation into Markov processes even though I have almost no recollection of how to work with them. It certainly looks to be more powerful than my way. Obviously my way and mathematician's way are mathematically identical although I flatter myself that my explanation may be slightly more intuitive. Of course I could not give an explanation in Nepali at all. Denis, mon ami, if you still are not convinced, I suggest that you try to give the reasoning that leads to your result.
 June 16th, 2017, 04:33 PM #17 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 I was kinda kidding...didn't want to admit I had badly programmed my simulation runs... Yes yes...I'm heading for the corner...
June 16th, 2017, 05:18 PM   #18
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Quote:
 Originally Posted by Denis I was kinda kidding...didn't want to admit I had badly programmed my simulation runs... Yes yes...I'm heading for the corner...
I probably will be joining you there soon enough.

 June 18th, 2017, 10:27 AM #19 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,908 Thanks: 716 To: Probability Prof. Jeff From: student Denis If problem changed to 1 to 9 (instead of 1 to 255) and to 4 picks (instead of 24), everything else remaining same, then: total combos = 9^4 combos with 1 pair + 2 diff. numbers: 3024 probability: 3024/9^4 = .460905... OR: 1 * 8/9 * 7/9 * 3/9 * 2 = .460905... Do I visit the corner again?
June 19th, 2017, 10:58 PM   #20
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Respectfully to All,

I think that the answer to the problem as stated (exactly one matching pair with no repeats) is ~0.38924, or precisely:

$\displaystyle \large \frac{255!}{232! \cdot 255^{24}} \cdot \frac{24!}{2! \cdot 22!}$

Notice that this is exactly 12 times the agreed upon value, given approximately as 0.0324, because:

$\displaystyle \large \frac{24!}{2! \cdot 22!} = 12 \cdot 23$

Please permit me to:
1) Show an inconsistency in the analysis of the agreed upon result.
2) Explain my concept of the solution.
3) Demonstrate the concept using what I think is the simplest example.
4) Provide a little background as to my interest in this.

So, for 1):

Quote:
 How many ways can I draw 23 differently numbered cards from that same set with replacement? Answer: $\displaystyle \large \prod_{j=1}^{23}(256 - j).$ Do you agree? How many ways can I draw one of 23 specific numbers from that set? Answer: 23.
So we have, I think:

$\displaystyle 255 \cdot 254 \cdot 253 \cdot \ldots \cdot 234 \cdot 233 \cdot 23$

If that is the case, then in the simplified example given earlier in the thread:

Quote:
 Two numbers the same (be careful: the different number can come first, second or third) = $\displaystyle 6∗5∗3=90.$
Why isn't that 3 a 2?

By the logic given in the first quote above, we can match 2 specific numbers here, not 3. But as is cautioned, it is not the number of numbers that is important, it is the number of places that is important. This appears to be inconsistent to me.

So for 2):

I would word it differently: It is the number of combinations of 3 numbers taken two at a time (the two types are numbers that match and the numbers that do not match):

$\displaystyle \large \frac{3!}{2! \cdot 1!}=3$

So to explain my solution, for the first number we have our choice of 255. Then let's match it 'right off the bat' with the second number so we have a choice of 1, then we can't match any more so we count down from there:

$\displaystyle \large 255 \cdot 1 \cdot 254 \cdot 253 \cdot \ldots \cdot 234 \cdot 233$

Obviously, we still have two types of numbers, the matching ones, and the non-matching ones, which gives the multiplication by:

$\displaystyle \large \frac{24!}{2! \cdot 22!} = 12 \cdot 23$

for the various combinations.

For 3) let's try to simplify things as much as we can. Let's have 3 symbols and take 4 of the symbols and calculate how many combinations exactly two of the symbols appear in.

I think that it is:

$\displaystyle \large 3 \cdot 1 \cdot 2 \cdot 1 \cdot \frac{4!}{2! \cdot 2!}=36$

If I am not mistaken, by the method of calculation as described in the thread the number is:

$\displaystyle \large 3 \cdot 2 \cdot 1 \cdot 3=18$

So let's just list them using symbols 0,1,2. Taking 2 as the repeated symbol, in ascending order we have:

0122
0212
0221
1022
1202
1220
2012
2021
2102
2120
2201
2210

So we have 12 combinations with the 2 repeated. We can have the 0 and 1 repeated as well so we have 3*12=36 (not 18 ).

For 4), I became interested in this type of thing because at my place of employment, in order to log in remotely, we have to enter a code that is chosen at random, 6 each of the digits 0-9.

I noticed that it was relatively rare to have no repeated digits so I tried to figure out the odds of getting exactly two repeating digits (among other things). I wrote programs in Ruby (which I will gladly share but I doubt anyone uses Ruby here) to run random numbers and looking at the percentage, and also to calculate exactly by running through all of the combinations and looking at the percentage.

I am pretty sure that I used reasoning similar to that given earlier in the thread but the calculated results did not match the programs' outputs until I used the methods given above.

Caveat: My 'exactly two matching numbers' routine may be flawed but for base 3 with 4 symbols taken, it does give 36, in agreement with the result above. For base 10 and 6 numbers taken it gives 453600 which agrees with:

$\displaystyle \large 10 \cdot 1 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot \frac{6!}{2! \cdot 4!}$

My random number calculating program calculates ~0.39 for the stated problem, in agreement with the calculation (and Denis' initial result, which I think is correct).

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HomeCSIRUGC NETThe probability that a ticketless traveler is caught during a trip is...

# The probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips , the probability that he/she will be caught during at least one of the trips is:

## Question:

The probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips , the probability that he/she will be caught during at least one of the trips is:
1. $1-(0.9)^4$
2. $(1-0.9)^4$
3. $1-(1-0.9)^4$
4. $(0.9)^4$

The probability that a ticketless traveler is caught during a trip is 0.1.
$therefore p=0.1$ and $q=1-p=1-0.1=0.9$
Since, traveler makes 4 trips, $n=4$.
Let X be the random variable that the traveler is caught during a trip.
Hence, $X leadsto B(n=4, p=0.1)$ and $X=0, 1, 2, 3, 4$.
$therefore P[X=x]=$ $^4C_x$ $(0.1)^x(0.9)^{4-x}$

$therefore$ the probability that he/she will be caught during at least one of the trips
=1-{probability that he/she will be caught during at most one of the trips}.
$=1- P[X=0]$
$=1-$ $^4C_0$ $(0.1)^0(0.9)^4$
$=1-(0.9)^4$
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# Probabilities of archers A and B hitting the bull's eye

1. Jan 10, 2013

### jinhuit95

1. The problem statement, all variables and given/known data
Two archers A and B take turns to shoot, with archer A taking the first shot. The probabilities of archers A and B hitting the bull's eye is 1/6 and 1/5 respectively. Show that the probability of archer A hitting the bull's eye first is 1/2.

2. Relevant equations

3. The attempt at a solution
Well, i thought about drawing tree diagram but the problem is I have no idea when to stop because I don't know when he will hit.

2. Jan 10, 2013

### dx

Re: Probability

First, what are the possible ways for this to happen?

A could hit the bulls eye on his first attempt. Or, A misses, B misses and then A hits the bulls eye. Or, A misses, B misses, A misses, B misses and then A hits the bulls eye... and so on.

Find the probability for each of these ways (for A to hit the bull's eye first), and add them up.

You will need to use the following:

1 + a2 + a3 + ... = 1/(1-a) for |a| < 1

3. Jan 10, 2013

### jinhuit95

Re: Probability

the problem is, if he hits on the first attempt that's 1/6. so If i go on, it will be 1/6 + ( 5/6 * 4/5 * 5/6 * 4/5 * 5/6 * 4/5....) I have no idea when to stop. Moreover, the product of probability he misses will be come smaller and smaller and that number + 1/6 will not give me half i think.

4. Jan 10, 2013

### dx

Re: Probability

Let's take the case when A gets it on his second try: A misses, B misses, and then A hits. What is the probability for this?

5. Jan 10, 2013

### jinhuit95

Re: Probability

that will be 5/6 * 4/5 * 1/6 right?

6. Jan 10, 2013

### jinhuit95

Re: Probability

I'm sorry but what's this?
'1 + a2 + a3 + ... = 1/(1-a) for |a| < 1'

7. Jan 10, 2013

### dx

Re: Probability

Yes, correct.

Now what is the probability that A gets it on his 3rd try? (A misses, B misses, A misses, B misses, A hits bull's eye)

Ignore that for the moment. We'll get to it soon.

8. Jan 10, 2013

### jinhuit95

Re: Probability

5/6* 4/5 * 5/6 * 4/5 * 1/6

9. Jan 10, 2013

### dx

Re: Probability

Correct.

So do you see the pattern?

First try: 1/6

Second try: (1/6)(4/6)

Third try: (1/6)(4/6)2

Fourth try: (1/6)(4/6)3

...

What is the probability that he gets in on the n'th try? Each time you are simply multiplying the previous one by (5/6)(4/5) = 4/6

10. Jan 10, 2013

### jinhuit95

Re: Probability

1/6 * 4/6 ^ n-1!

11. Jan 10, 2013

### dx

Re: Probability

Exactly.

Now you simply add them up. Let a = 4/6

(1/6) + (1/6)a + (1/6)a2 + (1/6)a3...

which is (1/6)(1 + a + a2 + a3 + a4...)

This is where you use the formula I mentioned before.

12. Jan 10, 2013

### jinhuit95

Re: Probability

Got it! Thanks alot!!

13. Jan 10, 2013

### dx

Re: Probability

No problem.

14. Jan 10, 2013

### Ray Vickson

Re: Probability

Here is another approach: let x = the probability that A will win, given that A is just about to shoot.

If A hits the target on his first shot, he wins; the probability of this is 1/6. If A misses his first shot (prob. = 5/6), then for A to win it must be the case that B misses his first shot (prob = 4/5); at that point, A is just about to shoot and we are back to the starting point (win prob. = x at that point). So, altogether we have
$$x = \frac{1}{6} + \frac{5}{6} \frac{4}{5}\, x = \frac{1}{6} + \frac{2}{3} x.$$
Solving, we get x = 1/2.

15. Jan 10, 2013

### CAF123

Re: Probability

@Ray
Is there a reason why doing: $$x = \frac{1}{6} + \frac{2}{3}x + \frac{5}{6} \frac{4}{5} \frac{5}{6} \frac{4}{5}x,$$ and solving is incorrect? (I am just considering the possibility of both A and B losing until A hits on his third go).

16. Jan 10, 2013

### Joffan

Re: Probability

My friend Kim and I had a similar competition, but we're much worse. I am slightly worse than Kim, so I went first. On any one shot, my chance of hitting the bull is 1/39, and Kim's chance is 1/38.

Show that, once again, this is an even competition - my chance of getting the bull's eye first is 1/2.

17. Jan 10, 2013

### CAF123

Re: Probability

Are you stuck? If so, just use exactly the same method as you used to solve the first one. Alternatively, say you win on your $i$th shot. This means you (or Kim) cannot have hit the bull on the previous $i-1$ attempts.
What is the prob of you not hitting the bull on the $i-1$ attempts?
What is the prob of Kim not hitting the bull on her $i-1$ attempts?
What is the prob of you hitting on the $i$th trial given that you and Kim failed on the $i-1$ trials?
By the independence of trials you can then multiply these together and sum.
(This is really the same method by dx just reworded a bit differently)

18. Jan 10, 2013

### haruspex

Re: Probability

Just use Ray's method again. It's much the simplest way to solve these problems. Did you understand Ray's method?

19. Jan 10, 2013

### Joffan

Re: Probability

Sorry for the misunderstanding guys. I was giving an extension question to illustrate the reasoning; I'm not stuck.

20. Jan 10, 2013

### Ray Vickson

Re: Probability

We can see it is incorrect by solving it and getting the wrong answer (x = -3/2 instead of x = 1/2). To understand WHY it is wrong, go back and look at the definition of x and the argument I gave for the equation satisfied by x. The point is that x already includes all the possibilities of winning on the first, third, fifth, .... go.

However, it would be correct to write
$$x = \frac{1}{6} + \frac{2}{3}\frac{1}{6} + \frac{2}{3}\frac{2}{3}\,x$$
for similar reasons. This equation does have the correct solution x = 1/2.

Last edited: Jan 10, 2013


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# A farmer has an apple orchard consisting of Fuji and Gala

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A farmer has an apple orchard consisting of Fuji and Gala  [#permalink]

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Updated on: 12 Aug 2012, 07:01
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Difficulty:

65% (hard)

Question Stats:

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A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

A. 22
B. 33
C. 55
D. 77
E. 88

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Originally posted by iamseer on 26 Apr 2010, 15:13.
Last edited by Bunuel on 12 Aug 2012, 07:01, edited 1 time in total.
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

### Show Tags

02 Apr 2011, 07:04
3
3
jay121 wrote:
Please tell me how should I calculate 187*0.85=220 within 30 seconds?

Is there any specific trick for that?

Even if i split it up and say: 0.1x +0.75x = 187 I have difficulties to solve it fast? Any tipps?

Thanks

10% are cross pollinated and 75% are pure Fuji so 85% are Fuji which gives you

$$(\frac{85}{100})*x = 187$$

Now it is obvious that number of trees has to be an integer so 85 and 187 need to have come common factor.
187 isn't divisible by 2 (not even), by 3(1+8+7 = 16 so not divisible by 3), by 7 (since 7*2 = 14, you have 47 left which will not go by 7) but it is divisible by 11 (11*1 = 11 and 11*7 = 77). So you split 187 into 11*17.
Now 85 is 17*5.

Now the equation becomes:
$$(\frac{17*5}{100})*x = 11*17$$
17 gets canceled and 5 gets canceled with 100 leaving a 20. So you get x = 220
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

### Show Tags

26 Apr 2010, 15:17
3
1
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

Let the total trees be x

3/4 are pure Fuji = 3x/4

10% cross pollinated = x/10

now The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187

3x/4 + x/10 = 187
solve this x = 220

220-187 = 33 are the pure Gala trees.
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

### Show Tags

Updated on: 26 Apr 2010, 15:21
1
2
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

Let f = pure fuji , g = pure gala and c - cross pollinated.

c = 10% of x where x is total trees.
c = .1x

also 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220

220 - 187 = pure gala = 33.

PS: While posting the questions please post options and source.
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Originally posted by gurpreetsingh on 26 Apr 2010, 15:19.
Last edited by gurpreetsingh on 26 Apr 2010, 15:21, edited 1 time in total.
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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26 Apr 2010, 15:21
gurpreetsingh wrote:
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

Let f = pure fuji , g = pure gala and c - cross pollinated.

c = 10% of x where x is total trees.
c = .1x

also 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220

220 - 187 = pure gala = 33.

Hi shouldnt it be 33, since 187 includes the fuji and the ones that cross pollinated???
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

### Show Tags

26 Apr 2010, 15:23
Silvers wrote:
gurpreetsingh wrote:
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

Let f = pure fuji , g = pure gala and c - cross pollinated.

c = 10% of x where x is total trees.
c = .1x

also 3x/4 = f and c+f = 187 => .1x + 3/4x = 187 => x = 220

220 - 187 = pure gala = 33.

Hi shouldnt it be 33, since 187 includes the fuji and the ones that cross pollinated???

Its 33 only check again what you have quoted.
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04 Jul 2010, 11:16
1
First - I want to point out the fact that the Google ad on this particular thread is an advertisement that says, "Apple trees are on sale now!"
Classic.

I'm not sure how to do this without simply setting up the formulas and hacking your way through the algebra. I'm guessing from your post ("need to understand the easiest way to solve this one") that you don't want this method.

Anyone else?
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04 Jul 2010, 11:52
yup....no need for the 4 equations method
f+g= T
0.1T = c
0.75T = F
f+c= 187
you can figure out the rest.....i hope there is a simpler way of doing it....
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05 Jul 2010, 13:10
2
Gala apples - G
Fuji apples - F
C - Cross Pollinated apples
Total apples - X
Given:
F = 3/4X
X = G + F + cross pollinated
Fuji + Cross pollinated (10 % of all apples) = 187

Solution:
3/4X + 1/10X = 187
Hence X = 220
X = G + F + cross pollinated
220 = G + 187
Hence G = 33.
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05 Jul 2010, 14:05
zisis wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

(A) 22
(B) 33
(C) 55
(D) 77
(E) 88

OA:

please explain method.....need to understand the easiest way to solve this one

I got confused with the highlighted part; how would the question be framed if the part in bold where to mean
pure fuji plus cross-pollinated fuji(not including cross-pollinated Gala)? Am I reading too much meaning to the question?
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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01 Apr 2011, 21:04
Please tell me how should I calculate 187*0.85=220 within 30 seconds?

Is there any specific trick for that?

Even if i split it up and say: 0.1x +0.75x = 187 I have difficulties to solve it fast? Any tipps?

Thanks
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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01 Apr 2011, 21:35
1
Let x be the number of apple trees

0.1x of apple trees are cross-pollinated

3x/4 = Fuji

3x/4 + 0.1x = 187

=> 0.85x = 187

=> 0.05x = 11

=> x = 1100/5 = 220

Now x/4 - 0.1x = Pure Gala

= 55 - 22 = 33
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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02 Apr 2011, 13:24
Lets assume F stands for pure Fuji, G for pure Gala , CP for cross pollinated and T for total trees.

F + CP = 187

=> 3T/4 + 10T/100 = 187 = > T = 220

T = F+G+CP = > G = 33
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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02 Apr 2011, 13:48
iamseer wrote:
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

10% of his trees cross pollinated i.e. Cross Pollinated = 0.1*Total
The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187 i.e. Pure Fuji+ Cross Pollinated=187
3/4 of all his trees are pure Fuji i.e. Pure Fuji=0.75*Total

Pure Fuji+ Cross Pollinated=0.75*Total+0.1*Total=0.85*Total

0.85*Total=187
Total=187/0.85

Pure Gala = Total - (Pure Fuji+ Cross Pollinated) = 187/0.85-187=187((1/0.85)-1)=187(0.15/0.85)=187(3/17)=11*3=33

Ans: 33 Pure Gala Apples
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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01 May 2011, 23:35
0.1x + 0.75x = 187
x = 220

number of G trees = 220-187 = 33
Hence B
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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12 Aug 2012, 05:53
Hello,

I got this question today on my MGMAT CAT and tried first to solve it through a chart for overlapping sets, but then realized that this is not possible. I solved the question afterwards as in the posts here equation system). How can I identify that the chart won't work here?

Thanks!
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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22 Sep 2012, 14:17
rocketscience wrote:
Hello,

I got this question today on my MGMAT CAT and tried first to solve it through a chart for overlapping sets, but then realized that this is not possible. I solved the question afterwards as in the posts here equation system). How can I identify that the chart won't work here?

Thanks!

2X2 works as well. You just need to set "neither" as 0 because apples have to be either G or F.

-------F -----nF
G----.1x-----y
nG---.75x---0
------187----?---x

From here solve for x and y will be the difference between x and 187.
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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04 Oct 2012, 15:48
this is the explanatio i understood

can anyone explain y it should be 55-22

Thank

subhashghosh wrote:
Let x be the number of apple trees

0.1x of apple trees are cross-pollinated

3x/4 = Fuji

3x/4 + 0.1x = 187

=> 0.85x = 187

=> 0.05x = 11

=> x = 1100/5 = 220

Now x/4 - 0.1x = Pure Gala

= 55 - 22 = 33
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Re: A farmer has an apple orchard: MGMAT, PS  [#permalink]

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04 Oct 2012, 21:16
venmic wrote:
this is the explanatio i understood

can anyone explain y it should be 55-22

Thank

subhashghosh wrote:
Let x be the number of apple trees

0.1x of apple trees are cross-pollinated

3x/4 = Fuji

3x/4 + 0.1x = 187

=> 0.85x = 187

=> 0.05x = 11

=> x = 1100/5 = 220

Now x/4 - 0.1x = Pure Gala

= 55 - 22 = 33

3/4 (i.e. 75%) are pure Fuji and 10% are cross so 15% are pure Gala.

So once you get x, you can calculate pure Gala as 15% of x = 15/100 * 220 = 33

or you can say Pure Gala = 25% of x (Gala apples) - 10% of x (cross) = 55 - 22 = 33 (as done above)
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Re: A farmer has an apple orchard consisting of Fuji and Gala  [#permalink]

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05 Oct 2012, 00:45
2
1
A farmer has an apple orchard consisting of Fuji and Gala apple trees. Due to high winds this year 10% of his trees cross pollinated. The number of his trees that are pure Fuji plus the cross-pollinated ones totals 187, while 3/4 of all his trees are pure Fuji. How many of his trees are pure Gala?

A. 22
B. 33
C. 55
D. 77
E. 88

THE QUICK METHOD...

Fuji + Cross = 187
10% are cross
75% are Fuji

so 85% = 187

We want to know what the 15% is

Divide our percent by 10, 8.5% = 18.7
Double it, 17% = 38

We need 15 percent and it is pretty obvious 33 fits the bill
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Re: A farmer has an apple orchard consisting of Fuji and Gala &nbs [#permalink] 05 Oct 2012, 00:45

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