# Probability of tossing a fair coin with at least $k$ consecutive heads

Tossing a fair coin for $N$ times and we get a result series as $HTHTHHTT\dots~$, Here '$H$' denotes 'head' and '$T$' denotes 'tail' for a specific tossing each time.

What is the probability that the length of the longest streak of consecutive heads is greater than or equal to $k$? (that is we have a $HHHH\dots~$, which is the substring of our tossing result, and whose length is greater than or equal to $k$)

I came up with a recursive solution (though not quite sure), but cannot find a closed form solution.

Here is my solution.

Denote $P(N,k)$ as the probability for tossing the coin $N$ times, and the longest continuous heads is greater or equal than $k$. Then (For $N>k$)

$$P(N,k)=P(N-1,k)+\Big(1-P(N-k-1,k)\Big)\left(\frac{1}{2}\right)^{k+1}$$

• Your example sequence reminded me of this brilliant Simpsons scene :-) – joriki Nov 10 '12 at 8:58
• I'm getting $(N-k+1)/2^k$ for the closed form. Would you like for me to post my solution, or do you want to think about it some more before I spoil the beans? – Braindead Nov 10 '12 at 8:59
• @Braindead: That can't be right; it's $\gt1$ for small $k$. – joriki Nov 10 '12 at 9:00
• Duh, you are right. I overcounted. – Braindead Nov 10 '12 at 9:08
• I tried to clarify some of the formulations; please check whether I preserved the intended meaning. In particular, I assumed that you had merely accidentally written "greater" once instead of "greater or equal". – joriki Nov 10 '12 at 9:12

Your recurrence relation is correct. I don't think you can do much better than that for general $k$, but you can find a closed form for specific values of $k$. For the first non-trivial value of $k$, the recurrence relation is

$$p_n=p_{n-1}+(1-p_{n-3})/8\;.$$

With $p_n=1+\lambda^n$, the characteristic equation becomes $\lambda^3-\lambda^2+1/8=0$. One solution, $\lambda=1/2$, can be guessed, and then factoring yields $(\lambda-1/2)(\lambda^2-\lambda/2-1/4)$ with the further solutions $\lambda=(1\pm\sqrt5)/4$. Thus the general solution is

$$p_n=1+c_1\left(\frac12\right)^n+c_2\left(\frac{1+\sqrt5}4\right)^n+c_3\left(\frac{1-\sqrt5}4\right)^n\;.$$

The initial conditions $p_0=0$, $p_1=0$, $p_2=1/4$ determine $c_1=0$, $c_2=-(1+3/\sqrt5)/2$ and $c_3=-(1-3/\sqrt5)/2$, so the probability is

\begin{align} p_n &=1-\frac{1+3/\sqrt5}2\left(\frac{1+\sqrt5}4\right)^n-\frac{1-3/\sqrt5}2\left(\frac{1-\sqrt5}4\right)^n\\ &=1-\frac4{\sqrt5}\left(\left(\frac{1+\sqrt5}4\right)^{n+2}-\left(\frac{1-\sqrt5}4\right)^{n+2}\right)\;. \end{align}

Thus, for large $n$ the probability approaches $1$ geometrically with ratio $(1+\sqrt5)/4\approx0.809$.

• I've added an explicit formula for the $P(N,k)$ which might be of interest to you. Note, that your $p_n=1-\frac{1}{2^n}F_{n+2}$ with $F_n$ the Fibonacci numbers. Regards, – Markus Scheuer Jan 18 '16 at 21:40

We can derive an explicit formula of the probability $P(N,k)$ based upon the Goulden-Jackson Cluster Method.

We consider the set of words $\mathcal{V}^{\star}$ of length $N\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the bad word $\underbrace{HH\ldots H}_{k \text{ elements }}=:H^k$ which is not allowed to be part of the words we are looking for. We derive a function $f_k(s)$ with the coefficient of $s^N$ being the number of wanted words of length $N$.

The wanted probability $P(N,k)$ can then be written as \begin{align*} P(N,k)=1-\frac{1}{2^N}[s^N]f_k(s) \end{align*}

According to the paper (p.7) from Goulden and Jackson the generating function $f_k(s)$ is \begin{align*} f_k(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\tag{1} \end{align*} with $d=|\mathcal{V}|=2$, the size of the alphabet and with $\mathcal{C}$ the weight-numerator with \begin{align*} \text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[H^k]) \end{align*} We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[H^k])&=-\frac{s^k}{1+s+\cdots+s^{k-1}}=-\frac{s^k(1-s)}{1-s^k} \end{align*}

We obtain the generating function $f(s)$ for the words built from $\{H,T\}$ which don't contain the substring $H^k$ \begin{align*} f_k(s)&=\frac{1}{1-2s+\frac{s^k(1-s)}{1-s^k}}\\ &=\frac{1-s^k}{1-2s+s^{k+1}}\tag{2}\\ \end{align*}

Note: For $k=2$ we obtain \begin{align*} f_2(s)&=\frac{1-s^2}{1-2s+s^{3}}\\ &=1+2s+3s^2+5s^3+8s^4+13s^5+21s^6+34s^7+\mathcal{O}(s^8) \end{align*}

The coefficients of $f_2(s)$ are a shifted variant of the Fibonacci numbers stored as A000045 in OEIS.

Note: For $k=3$ we obtain \begin{align*} f_3(s)&=\frac{1-s^3}{1-2s+s^{4}}\\ &=1+2s+4s^2+7s^3+13s^4+24s^5+44s^6+81s^7+\mathcal{O}(s^8) \end{align*}

The coefficients of $f_3(s)$ are a shifted variant of the so-called Tribonacci numbers stored as A000073 in OEIS.

We use the series representation of $f_k(s)$ in (2) to derive an explicit formula of the coefficients.

\begin{align*} [s^N]f(s)&=[s^N](1-s^k)\sum_{m=0}^{\infty}(2s-s^{k+1})^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m(2-s^k)^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\\ &=([s^N]-[s^{N-k}])\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{3}\\ &=\sum_{m=0}^{N}([s^{N-m}]-[s^{N-k-m}])\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{4}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ &\qquad-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N-k}\binom{m}{\frac{N-m}{k}-1}(-1)^{\frac{N-m}{k}-1}2^{m-\frac{N-m}{k}+1}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\tag{5}\\ &\qquad+\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ \end{align*}

Comment:

• In (3) we use the linearity of the coefficient of operator and $[s^N]s^mf(s)=[s^{N-m}]f(s)$

• In (4) we change the limit of the left hand sum from $\infty$ to $N$ according to the maximum coefficient $[s^N]$. According to the factors $s^{kj}$ we consider in the following only summands with $m\equiv N(\bmod k)$

• In (5) we reorganise the sums from the line above by extracting from the left hand sum the first summand and shifting in the right hand side the index by one and putting both sums together.

We conclude: An explicit representation of the probability $P(N,k)$ $(n\geq 0)$ is according to (5) \begin{align*} P(N,k)&=1-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}}\\ &\qquad-\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}} \end{align*}

A bit more practical focus than previous answers:

Main idea:

1. We build a "staircase":

2. $p$ chance to get to next step, $1-p$ chance to fall down and have to restart (or "reflip"). Except for the highest step where we always stay (whenever we manage to get there).

3. We need $k+1$ steps to "remember" where on the stairs we are for $k$ flips in a row.

We can use this to build a matrix for a Markov chain:

We can build a block matrix:$$\frac{1}{2}\left[\begin{array}{cc}2&1&\bf 0^T\\\bf 0&\bf0&\bf I_k\\0&1&\bf 1^T \end{array}\right]$$

(For the special case $k = 5$). We build a stochastic matrix:

$${\bf P} = \frac{1}{2}\left[\begin{array}{cccccc}2&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\0&1&1&1&1&1\end{array}\right]$$

Now our answer will simply be $$[1,0,{\bf 0}]\, {\bf P}^k\, [{\bf 0},0,1]^T$$

Which is a scalar product using ${\bf P}^k$ as a gramian matrix.

Now to calculate this in practice one could probably in general (for general $p$) benefit from a canonical transformation of $\bf P$, but it's doubtable in this case as the matrix is already sparse with literally only sums and permuations and bit shifts required on the vector elements.

Feller considers this problem in section XIII.7 of An Introduction to Probability Theory and Its Applications, Volume 1, Third Edition. He shows that the probability of having no run of length $k$ in $N$ throws is asymptotic to $$\frac{1-(1/2)\;x}{(k + 1 - kx)\; (1/2)} \cdot \frac{1}{x^{N+1}}$$ where $x$ is the least positive root of $$1 - x +(1/2)^{k+1} x^{k+1} = 0$$

(I have changed Feller's notation to agree with the problem statement and have only considered the case of a fair coin; Feller considers the more general case of a biased coin. For more information see equation 7.11, p. 325 in the referenced document.)


# Can an open set be covered by proper open subsets?

Let $(X, \tau)$ be a topological space, let $U\in\tau$. An open cover of $U$ is a set $\{U_i \ |\ i\in I\}$ (of open sets $U_i$) whose union $\bigcup U_i$ contains $U$.

If $U\subsetneq X$, then $U$ admits an open covering by open sets $\{U_i \ |\ i\in I\}$, where an arbitrary open set $U_i$ may not be a subset of $U$. (I think.)

Does $U$ admit an open covering by open sets $\{U_i \ |\ i\in I\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets "inside" of $U$?

If the answer is no, then what are the open sets $U$ that admit such "internal" coverings?

Is the answer any different if we replace $U$ by an arbitrary closed set?

• Have you thought about the integers? – Steve D May 15 '17 at 21:38
• For many topological spaces you can do this. For example, any open subset of $\mathbb R^n$ is a union of open balls, which can be made proper. However, @SteveD's hint shows that you can't always do this. – Cheerful Parsnip May 15 '17 at 22:20
• If $X$ is a $T_1$ space and if $U$ is open and has more than $1$ member then $\{U$ \ $\{p\}: p\in U\}$ is an open cover of $U$ by open proper subsets of $U.$ ....If $\tau=\{X,\phi\}$ then with U=X, no proper non-empty subset of $U$ is open.... Another example is Sierpinski space $S=\{0,1\}$ with $\tau=\{S,\phi, \{0\}\}$. – DanielWainfleet May 15 '17 at 23:57
• If U is open, of course. If U is open every point is an interior point so for every point x in U there is an open neighborhood of x completely in U. Call this N_x, then $\cup_{x\in U} Nx$ is an open cover. If U is not open then of course not, If U is not open there is a point that is not an interior point and any open set containing it must contain points not in U (else it'd be an interior point). So all open covers contain points not in U. – fleablood May 16 '17 at 1:25
• @fleablood This is the argument I was going to give, but (as others pointed out to me) it only goes through if the space is at least $T_1$. – Austin Mohr May 16 '17 at 1:31

A space is called $T_1$ if, for any two points in the space, each has an open neighborhood that misses the other. This is one of many separation axioms that measure how strongly we can separate points in the space. For some mathematicians, topological spaces aren't even worth considering until they are at least $T_2$ (Hausdorff), which is even stronger. (This is to say that all but the most pathological topologies are at least $T_1$.)

We can construct a cover of the type you describe if the space is at least $T_1$ and $U$ contains more than one point. For each $x \in U$, construct an open neighborhood $U_x$ as follows:

1. Choose any $y \in U$ that is distinct from $x$.
2. Appeal to the $T_1$ property to get an open set $G_x$ that contains $x$ but does not contain $y$.
3. Set $U_x = G_x \cap U$, so that $U_x$ is a proper open subset of $U$ containing $x$.

Finally, we can take $\{U_x \mid x \in U\}$ as an open cover of $U$ by proper open subsets.

• But why is it guaranteed that the "first" $y$ we chose is an element of some element of $\{ U_x \ | \ x\in U \}$ ? Moreover, why is every such $y$ eventually in some of the cover elements? Because, by construction, $y$ is not in $U_x$ – étale-cohomology May 17 '17 at 5:52
• Let $y_0$ be the "first $y$" chosen in this process. Eventually we will need to construct the set $U_{y_0}$, which must contain $y_0$. Similarly, any $y \in U$ is an element of $U_y$ and therefore covered. – Austin Mohr May 17 '17 at 8:36
• Got it. Thank you. I wonder if it's possible to prove that $T_1$ is the weakest possible hypothesis that allows existence of these covers... – étale-cohomology May 18 '17 at 22:38
• One can push my argument through with something weaker (but I don't know if it has a name): For any open set $U$ and any point $x$ of $U$, there is an open neighborhood $V$ of $x$ that is a strict subset of $U$. This property is not as strong as $T_1$, since it still allows for the existence $x$ and $y$ that cannot be separated. However, you could take the $V$ in this property as the $U_x$ in my answer and it still get your desired refinement. This is property would be weakest possible, since otherwise the only way to cover $x$ is to use all of $U$. – Austin Mohr May 19 '17 at 1:01

It depends. Consider $\mathbb{R}$ with the usual topology and $U = (0,1)$, then we can write $(0,1) = \bigcup_{n=2}^\infty\left(\frac1n,1\right)$, for instance. Since every open set $U\subset\mathbb{R}$ is the union of intervals, it means that every open subset of $\mathbb{R}$ can be written as the union of proper open subsets of itself. It is easy to see that the same argument can be applied to general normed vector spaces.

However, if a singleton $\{x\}$ is an open set of your space then there are no proper open subsets of $\{x\}$ to unite.

Show that $K = [0,1]$ cannot be covered by open sets that are contained by $K$.

If $U$ is open, then $U$ can be covered by $\{U\}$. If $U$ is an open singleton, then there is no cover by proper subsets.

When $U$ is open, then for all $x$ in $U$, there is an open base set $U_x$ with $x$ in $U_x$ subset $U$. Often the base sets can be chosen to be proper subsets.


# Why is $x+e^{-x}>0$ for all $x \in \mathbb{R}$?

Denote $$f(x) = x + e^{-x}$$. Note that $$f(0) = 1$$ and $$f'(x) = 1-e^{-x}$$. That means $$\lim_{x\rightarrow -\infty} f'(x) = -\infty.$$ So if the rate of change of $$f(x)$$ keeps decreasing exponentially fast to negative infinity, why is $$\lim_{x\rightarrow -\infty} f(x) = +\infty$$?

• Addressing the title, for all $x, e^{-x}\ge1-x$ so $x+e^{-x}\ge1>0$ – J. W. Tanner Mar 21 at 4:45
• To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}\ge t+1$ for all real $t$. So $e^{-x}\ge -x+1> -x$ for all real $x$, as desired. – Minus One-Twelfth Mar 21 at 4:46
• Roughly speaking - if you are visualising $x$ going from $0$ to $-\infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards". – David Mar 21 at 4:49
• Thank you. That's what I was missing. Going backward. :( – Paichu Mar 21 at 4:50

You have got your directions mixed up.

When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.

Take

$$x < y < 0; \tag 1$$

then

$$f(y) - f(x) = \displaystyle \int_x^y f'(s)\; ds = \int_x^y (1 - e^{-s})\; ds; \tag 2$$

it is clear that, for any $$M < 0$$, there exists $$y_0 < 0$$ such that

$$s < y_0 \Longrightarrow 1 - e^{-s} < M; \tag 3$$

if we now choose

$$y < y_0, \tag 4$$

then

$$f(y) - f(x) = \displaystyle \int_x^y (1 - e^{-s})\; ds < \int_x^y M \; ds = M(y - x); \tag 5$$

we re-arrange this inequality:

$$f(x) - f(y) > -M(y - x) = M(x - y), \tag 6$$

$$f(x) > f(y) + M(x - y); \tag 7$$

we now fix $$y$$ and let $$x \to -\infty$$; then since $$M < 0$$ and $$x - y < 0$$ for $$x < y$$,

$$\displaystyle \lim_{x \to -\infty} f(y) + M(x - y) = \infty, \tag 8$$

and hence

$$\displaystyle \lim_{x \to -\infty}f(x) = \infty \tag 9$$

as well.

We note that (9) binds despite the fact that $$f'(x) < 0$$ for $$x < 0$$; though this derivative is negative, when $$x$$ decreases we are "walking back up the hill," as it were; as $$x$$ decreases, $$f(x)$$ increases.

Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show

$$x + e^{-x} > 0, \forall x \in \Bbb R; \tag{10}$$

for

$$x \ge 0, \tag{11}$$

$$e^{-x} > 0 \tag{12}$$

as well, hence we also have

$$x + e^{-x} > 0; \tag{13}$$

for

$$x < 0, \tag{14}$$

we may use the power series for $$e^{-x}$$:

$$e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3} + \ldots; \tag{15}$$

then

$$x + e^{-x} = 1 + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \ldots > 0, \tag{16}$$

since every term on the right is positive when $$x < 0$$. End of Note.

Note that $$f''(x) = e^{-x} >0$$ so $$f$$ is strictly convex.

Since $$f'(0) = 0$$, we see that $$f(x) \ge f(0) = 1$$ for all $$x$$.

Note that

• $$\forall x \in \mathbb{R} : x+e^{-x}>0 \Leftrightarrow \forall x \in \mathbb{R} :e^{-x}>-x \Leftrightarrow \forall x \in \mathbb{R} :\color{blue}{\boxed{e^{x}>x}}$$

The last inequality follows directly by Taylor: $$\color{blue}{e^x} = 1+x + \underbrace{\frac{e^{\xi}}{2}x^2}_{\geq 0} \color{blue}{> x}$$


# How can you express the next two consecutive odd numbers in terms of x?

I'm trying to express the next two consecutive odd numbers in terms of x. How do go about this?

Precisely, given a number $x$ I want two smallest consecutive odd numbers $m,n$ such that $x<m<n$.

• Be aware that you wrote two very different things in your question and your body. Are you trying to express the next two consecutive odd numbers in terms of x, or are you trying to express x in terms of the next two consecutive odd numbers? If $x$ is, say $5$, do you consider "the next two consecutive odd numbers" to be $5$ and $7$ or do you consider the next two consecutive odd numbers to be $7$ and $9$? – JMoravitz Oct 19 '15 at 2:32
• My error. I'm trying t express the next two consecutive odd numbers in terms of x. Edited to reflect. – user214824 Oct 19 '15 at 2:45
• That still doesn't answer the question of what exactly you mean by "next two consecutive odd numbers" is. If $5$ is your number, do you want $7$ and $9$? or do you want $5$ and $7$ (despite $5$ not being bigger than $5$). – JMoravitz Oct 19 '15 at 2:47
• @JMoravitz Yes, that is it. 7 and 9. – user214824 Oct 19 '15 at 2:48

Presumably $x$ is an integer. We have two cases: either $x$ is even or $x$ is odd.
In the case that $x$ is odd, then $x+2$ and $x+4$ will both be odd and will be the next two consecutive odd numbers.
In the case that $x$ is even, then $x+1$ and $x+3$ will both be odd and will be the next two consecutive odd numbers.
Let $x$ an integer, then the next two odd integers are $$2\times\Bigl\lfloor \frac{x+1}{2}\Bigr\rfloor + 1 \qquad\text{and}\qquad 2\times\Bigl\lfloor \frac{x+1}{2} \Bigr\rfloor + 3.$$
Indeed, if $x = 2n$ where $n\in\mathbb{Z}$, then $$2\lfloor (x+1)/2\rfloor + 1 = 2 \lfloor n + 1/2 \rfloor + 1 = 2n + 1$$ and if $x = 2n + 1$, then $$2\lfloor (x+1)/2 \rfloor + 1 = 2\lfloor n + 1 \rfloor + 1 = 2n + 3.$$
• Maybe doesn't you know the symbol $\lfloor \cdot \rfloor$. If $x$ is a real number, $\lfloor x \rfloor$ is the biggest integer less or equal to $x$. For example $\lfloor 2.7 \rfloor = 2$, $\lfloor -3.8 \rfloor = -4$ and $\lfloor x \rfloor = x$ for $x$ integer. – Éric Guirbal Oct 19 '15 at 3:01


# Is this theorem equivalent to Transfinite Recursion Theorem?

Let $$V$$ be the class of all sets, $$\operatorname{Ord}$$ be the class of all ordinals, and $$G:V\to V$$ be a class function.

Transfinite Recursion Theorem:

There exists a class function $$F:\operatorname{Ord}\to V$$ such that $$F(\alpha)=G(F\restriction \alpha)$$ for all $$\alpha\in\operatorname{Ord}$$.

While I'm able to prove that Transfinite Recursion Theorem implies the below theorem, I have tried but to no avail in proving that the theorem implies Transfinite Recursion Theorem.

I would like to ask if it is possible to prove that this theorem implies Transfinite Recursion Theorem.

Theorem:

Let $$G_1,G_2,G_3$$ be class functions from $$V$$ to $$V$$. There exists a class function $$F:\operatorname{Ord}\to V$$ such that

(1) $$F(0)=G_1(\emptyset)$$

(2) $$F(\alpha+1)=G_2(F(\alpha))$$ for all $$\alpha\in\operatorname{Ord}$$

(3) $$F(\alpha)=G_3(F\restriction\alpha)$$ for all limit $$\alpha\neq 0$$

• Do you know how to prove strong induction (on $\mathbb{N}$) from weak induction? It's the same idea. – Eric Wofsey Oct 25 '18 at 2:46
• Hi @EricWofsey! Yes, I do. I will try your suggestion. – LE Anh Dung Oct 25 '18 at 2:54

The trick is to not construct $$F$$ itself, but to construct the function $$H$$ which sends $$\alpha$$ to $$F\restriction \alpha$$. That way, at successor steps you have access to the entire history of the recursion so far and not just the last step.

In detail, given $$G:V\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: $$G_1(x)=\emptyset$$ $$G_2(x)=x\cup\{(\operatorname{dom}(x),G(x))\}$$ $$G_3(x)=\bigcup\operatorname{ran}(x)$$ By the theorem, we then get a function $$H:Ord\to V$$ such that $$H(0)=G_1(\emptyset)$$, $$H(\alpha+1)=G_2(H(\alpha))$$, and $$H(\alpha)=G_3(H\restriction\alpha)$$ for $$\alpha\neq 0$$ limit. It is then easy to prove by induction that $$H(\alpha)$$ is a function with domain $$\alpha$$ for each $$\alpha$$, with $$H(\alpha)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha$$. So, defining $$F=\bigcup\operatorname{ran}(H)$$, $$F$$ is a function on $$Ord$$ with $$F\restriction\alpha=H(\alpha)$$ for each $$\alpha$$. For each $$\alpha$$, we then have $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=G_2(F\restriction\alpha)(\alpha).$$ Since $$\operatorname{dom}(F\restriction\alpha)=\alpha$$, our definition of $$G_2$$ tells us that $$F(\alpha)=G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha),$$ as desired.

• I'm feeling confused. You have shown how to derive the Q's 2nd theorem from its first. But the proposer seems to want to see the 1st derived from the second, although it is obvious that if the second one holds, then, for a given $G$, we can simply let $G_1=G_2=G_3=G$. So I'm thinking that the proposer stated the Q backwards. – DanielWainfleet Oct 25 '18 at 3:49
• @DanielWainfleet: I am deriving the first theorem from the second. I start with a function $G$ as in the first theorem, then use the second theorem to get a function $H$ and then define $F$ from $H$ which verifies the conclusion of the first theorem. It does not work to simply define $G_1=G_2=G_3=G$ since $G_2$ takes as its input just $F(\alpha)$, not $F\restriction\alpha$. – Eric Wofsey Oct 25 '18 at 4:04
• Thank you so much! You make my day :) – LE Anh Dung Oct 25 '18 at 8:14
• Thank you for clearing up my confusion. – DanielWainfleet Oct 25 '18 at 21:26
• Hi! I'm now struggling with a seemingly similar problem like this one. I have tried to replicate your proof but failed at the end. If you don't mind, please have a look at math.stackexchange.com/questions/2970783/…. Thank you for your help! – LE Anh Dung Oct 27 '18 at 14:16

I fill in @Eric Wofsey's proof with detail and post it here. All credits are given to @Eric Wofsey.

Given $$G:V\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: \begin{align}&G_1(x)=\emptyset\text{ for all }x\\&G_2(x)=\begin{cases} x\cup\{(\operatorname{dom}(x),G(x))\}&\text{if }x\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}\\&G_3(x)=\begin{cases} \bigcup\operatorname{ran}(x)&\text{if }x\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}\end{align}

By The theorem, there is a class function $$H:\operatorname{Ord}\to V$$ such that $$H(0)=G_1(\emptyset)$$, $$H(\alpha+1)=G_2(H(\alpha))$$, and $$H(\alpha)=G_3(H\restriction\alpha)$$ for $$\alpha\neq 0$$ limit.

First, we prove by induction that $$H(\alpha)$$ is a function with domain $$\alpha$$ for all $$\alpha\in\operatorname{Ord}$$ and with $$H(\alpha)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha$$.

• $$H(0)=G_1(0)=\emptyset$$. Then the statement is trivially true for $$\alpha=0$$.

• Assume that the statement is true for $$\alpha$$. Then $$H(\alpha+1)=G_2(H(\alpha))=$$ $$H(\alpha)\cup\{(\operatorname{dom}(H(\alpha)),G(H(\alpha)))\}=H(\alpha)\cup\{(\alpha,G(H(\alpha)))\}$$. It follows that $$\operatorname{dom}(H(\alpha+1))=\operatorname{dom}(H(\alpha))\cup \{\alpha\}=\alpha\cup \{\alpha\}=\alpha+1$$. For $$\beta=\alpha$$, $$H(\alpha+1)\restriction \beta=H(\alpha+1)\restriction \alpha=H(\alpha)=H(\beta)$$. For $$\beta<\alpha$$, $$H(\alpha+1)\restriction \beta=$$ $$H(\alpha)\restriction \beta=H(\beta)$$. Thus $$H(\alpha+1)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha+1$$.

• Assume that the statement is true for all $$\beta<\alpha$$ where $$\alpha\neq\emptyset$$ is limit ordinal. Then $$H(\alpha)=G_3(H(\alpha))=\bigcup\operatorname{ran}(H(\alpha))=\bigcup\{H(\beta)\mid \beta<\alpha\}$$. For any $$\beta_1\le\beta_2<\alpha$$: $$H(\beta_2)\restriction\beta_1=H(\beta_1)$$and thus $$H(\beta_1)\subseteq H(\beta_2)$$. Then $$H(\alpha)=\bigcup\{H(\beta)\mid \beta<\alpha\}$$ is actually a function. It follows that $$\operatorname{dom}(H(\alpha))=\bigcup_{\beta<\alpha}\operatorname{dom}(H(\beta))=\bigcup_{\beta<\alpha}\beta=\alpha$$ since $$\alpha$$ is limit ordinal. Moreover, $$H(\alpha)\restriction \beta=\{(\gamma,H(\alpha)(\gamma))\mid \gamma<\beta\}=\{(\gamma,H(\gamma+1)(\gamma))\mid \gamma<\beta\}=$$ $$\{(\gamma,H(\beta)(\gamma))\mid \gamma<\beta\}=H(\beta)$$.

As a result, $$\forall\beta<\alpha:H(\alpha)\restriction \beta=H(\beta)$$ and thus $$\forall\beta<\alpha:H(\beta)\subsetneq H(\alpha)$$.

Next we define $$F=\bigcup\operatorname{ran}(H)$$. Then $$F=\bigcup\{H(\alpha)\mid \alpha\in\operatorname{Ord}\}$$ is a function with domain $$\operatorname{Ord}$$ and with $$F\restriction\alpha=\{F(\beta)\mid\beta<\alpha\}=\{H(\beta+1)(\beta)\mid\beta<\alpha\}=\{H(\alpha)(\beta)\mid\beta<\alpha\}=H(\alpha)$$ for all $$\alpha\in\operatorname{Ord}$$.

Since $$F\restriction\alpha=H(\alpha)$$ and $$\operatorname{dom}(H(\alpha))=\alpha$$, $$\operatorname{dom}(F\restriction\alpha)=\alpha$$. Then $$G_2(F\restriction\alpha)=(F\restriction\alpha)\cup \{(\alpha,G(F\restriction\alpha))\}$$ and thus $$G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha)$$.

For each $$\alpha\in\operatorname{Ord}$$, we have $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha)$$.

Update: I have found another way to define $$F$$

We define $$F$$ as follows $$F(\alpha):=H(\alpha+1)(\alpha)$$

Then $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=(H(\alpha)\cup\{(\operatorname{dom}(H(\alpha)),G(H(\alpha)))\})(\alpha)=(H(\alpha)\cup\{(\alpha,G(H(\alpha)))\})(\alpha)=G(H(\alpha))$$.

Moreover, $$H(\alpha)=\{(\beta,H(\alpha)(\beta))\mid\beta<\alpha\}=\{(\beta,H(\beta+1)(\beta))\mid\beta<\alpha\}=\{(\beta,F(\beta))\mid\beta<\alpha\}=F\restriction\alpha$$.

Thus $$F(\alpha)=G(H(\alpha))=G(F\restriction\alpha)$$.


# Envelope of Projectile Trajectories

For a given launch velocity $$v$$ and launch angle $$\theta$$, the trajectory of a projectile may be described by the standard formula $$y=x\tan\theta-\frac {gx^2}{2v^2}\sec^2\theta$$

For different values of $$\theta$$ what is the envelope of the different trajectories? Is it a parabola itself?

The standard solution to this "envelope of safety" problem is to state the formula as a quadratic in $$\tan\theta$$ and set the discriminant to zero. The resulting relationship between $$x,y$$ is the envelope.

This question is posted to see if there are other approaches to the solution.

Edit 1

Thanks for the nice solutions from Jack and Blue, received so far. From the solution of the envelope it can be worked out that the envelope itself corresponds to the right half of the trajectory of a projectile launched at $$(-\frac{v^2}{g^2},0)$$ at a launch angle $$\alpha=\frac{\pi}4$$ and a launch velocity $$V=v\sqrt2$$. This means that both vertical and horizontal components of the launch velociy are equal to $$v$$. It would be interesting to see if these conclusions can be inferred from the problem itself by inspection and without first solving it. If so, then this would form another solution.

• I think it is more like an ellipse. – Archis Welankar Oct 24 '15 at 10:05
• @ArchisWelankar - What makes you think that might be the case? – hypergeometric Oct 24 '15 at 10:13
• When the angle is very small like 10 then the curve becomes elongated and more like an ellipse than a parabola. – Archis Welankar Oct 24 '15 at 10:25
• @ArchisWelankar - 1. The curve or trajectory of the projectile is always a parabola (even when elongated, at small launch angles) as given by the formula. 2. An ellipse doesn't have to be elongated. 3. The question refers to the envelope of different trajectories and not any one particular trajectory. – hypergeometric Oct 24 '15 at 10:34
• "The standard solution to this "envelope of safety" problem is to state the formula as a quadratic in tanθ and set the discriminant to zero. " I don't understand this approach, can someone explain ? – Dat Sep 11 at 9:38

Yes it is. To find the envelope, we just have to find the intersections between two trajectories associated to two slightly different angles. If we solve $$x\tan\theta -\frac{gx^2}{2v^2}\sec^2(\theta) = x\tan(\theta+\varepsilon) -\frac{gx^2}{2v^2}\sec^2(\theta+\varepsilon)$$ we get $x=0$ or $$x=\frac{2v^2}{g}\cdot\frac{\tan(\theta)-\tan(\theta+\varepsilon)}{\sec^2(\theta)-\sec^2(\theta+\varepsilon)}$$ and by letting $\varepsilon\to 0$ we get $x=\frac{v^2}{g}\cot(\theta)$, from which $y=\frac{v^2}{2g}\left(2-\frac{1}{\sin^2(\theta)}\right)$.

It follows that the equation of the envelope is given by: $$y = \frac{v^2}{2g}\left(1-\left(\frac{gx}{v^2}\right)^2\right)=\frac{v^2}{2g}-\frac{g}{2 v^2}\,x^2$$ that clearly is a parabola with vertex in $\left(0,\frac{v^2}{2g}\right)$ through the points $\left(\pm\frac{v^2}{g} ,0\right)$.

We may notice that the envelope and the trajectory with $\theta=\frac{\pi}{4}$ are homothetic, and the dilation ratio is just $2$. The vertices of the trajectories lie on an ellipse that is tangent to the envelope parabola, with centre at $\left(0,\frac{v^2}{4g}\right)$, a vertex in the origin and a vertex at $\left(\frac{v^2}{2g},\frac{v^2}{4g}\right)$.

• Thank you for a beautiful and creative solution! And nice graphics! A big +1! – hypergeometric Oct 24 '15 at 13:29
• Please check out comments under Edit 1 in the question and see if you have any further thoughts. – hypergeometric Oct 24 '15 at 14:21

@Jack provides a very nice and intuitive derivation of the envelope as the points of intersection of infinitely-close members of the curve family. The Wikipedia "Envelope" entry provides this less-illuminating abstraction:

The envelope of the family [of curves parameterized by $t$ is] the set of points for which $$F(t, x, y) = \frac{\partial F}{\partial t}(t,x,y) = 0 \tag{\star}$$ for some value of $t$ [...].

In $(\star)$, $F$ is the function that, when set equal to $0$, defines each curve in the family. For the question at hand, we have (with parameter $\theta$ instead of $t$) $$F(\theta,x,y) = -y + x\tan\theta -\frac{g x^2}{2v^2}\sec^2\theta \tag{1}$$ Therefore, differentiating with respect to $\theta$ gives $$\frac{\partial F}{\partial \theta}(\theta,x,y) =x\sec^2\theta -\frac{g x^2}{2v^2}\cdot 2 \sec^2\theta\tan\theta = \frac{x\sec^2\theta}{v^2} ( v^2 - g x \tan\theta) \tag{2}$$ Solving $\partial F/\partial \theta = 0$ for $\theta$ (noting that $\sec\theta$ never vanishes) gives $$\tan\theta = \frac{v^2}{g x} \qquad\text{so that}\qquad \sec^2\theta = 1 + \tan^2\theta = \frac{g^2 x^2 + v^4}{g^2 x^2}$$

Substituting into $(1)$, and setting $F=0$, we have

$$y = x\;\frac{v^2}{gx} - \frac{g x^2}{2 v^2}\;\frac{g^2 x^2 + v^4}{g^2 x^2} = \frac{v^2}{2g} - \frac{g x^2}{2v^2} = \frac{v^4-g^2x^2}{2gv^2}$$

• Thanks for working out the solution from the article. Nice find! +1. – hypergeometric Oct 24 '15 at 13:55
• Please check out comments under Edit 1 in the question and see if you have any further thoughts. – hypergeometric Oct 24 '15 at 14:21

Here's another approach. Not as elegant as Jack's though. Perhaps similar to Blue's. (Edit: Revised slightly to use substitution of $H=v^2/2g$ instead of $\lambda=g/v^2$ for more convenient reference, per solution by Narasimham)

This approach maximises the value of $y$ for any given value of $x$.

For a given value of $x$, say $x=k$, \begin{align} y&=kT-\frac {k^2}{4H}(1+T^2)\qquad\text{where T=\tan\theta, H=\frac {v^2}{2g}}\\ \frac{dy}{dT}&=k-\frac {k^2}{4H}\cdot 2T=0\qquad\text{when T=\frac {2H}k}\\ \text{At T=\frac {2H}k}:\qquad\qquad y&=k\cdot\frac{2H}k-\frac{k^2} {4H}\left(1+\frac {4H^2}{k^2}\right) \color{lightgrey}{=2H-\frac {k^2}{4H}-H}\\ &=H-\frac {k^2}{4H} \end{align}

Hence the equation of the envelope is $$y=\frac {v^2}{2g}-\frac {gx^2}{2v^2}\quad\blacksquare$$

• This nicely exploits the fact that, in this particular case, we know that the envelope sits on top of the family of curves, and therefore consists of the points that maximize $y$ for a given $x$, over all $t$. (Of course don't always have such knowledge, but no matter.) Re-writing in terms of parameter $T$ is also a helpful simplification. And, to be explicit: the connection with the process in my answer is that the criterion for maximizing $y$ ---that is, setting $dy/dT = 0$--- corresponds directly to $\partial F/\partial \theta = 0$, because $y$ drops out of the latter equation. – Blue Oct 24 '15 at 21:38
• Thanks for your kind comment! – hypergeometric Oct 25 '15 at 5:57

I shall outline the method how we get the parabola. Usual notation

$$\ddot y = -g, \dot y = - g t + v \sin \theta , y =- g t^2/2 + v t \sin \theta\, +0 \tag {1}$$ $$\ddot x = 0, \dot x = v \cos \theta = const , x = v t \cos \theta +0 \tag{2}$$

Eliminating time $t$ between (1),(2) you got this parabola equation already.

Let $\tan \alpha = T$; Parabola equation in other words

$$y = x T - g/2 * ( x/ v \cos\theta)^2 = x T - ( g x^2/2 v^2) ( 1+T^2) \tag{3}$$

Differentiate partially with respect to $T$ and simplify, $T = v^2 / gx \tag{4}$

Eliminate $T$ between (3) and (4)

$$y = v^2/2g - g x^2 /(2 v^2) = H - x^2/(4H) \tag{5},$$

same as what was obtained before by Jack and Blue.

If you denote height reached by projectile on vertical firing $H = v^2/(2g) \tag{6}$ you would notice that the envelope is profiled exactly as a parabolic mirror with focus at gun delivery point, focal length is exactly H. Vertical force of gravity is acting like light :)..

The above procedure method is indicated by Blue in Wiki, is referred to as C-discriminant method to obtain envelopes and singular solutions.

Like what you said in your edit and I about mirror, they are ploys for remembering curves using similarities..

For the image I took values of $g=9.8 m/s^2 , v = 2 m/s$

The partial differentiation way and elimination is the right way to look at, perhaps not as what you said (standard solution.. way).

EDIT2:

The answer to your second question, i.e., to determine if it is going to be a parabola envelope without going through all of analysis... I can only reply with extended C-discriminant, strengthening the same result by another path.

$p$ discriminant method is also relevant, but I defer it, but best is to refer to differential calculus books of authors e.g., A.R. Forsythe.

I shall expand on the C-discriminant, wherein a two parameter system of equations variation of any of the two parameters leads to the same parabola envelope. This is in reply to your question , Why do you also plot partial concentric circles?

Well, they are circles alright, but not concentric circles. They expand and come down slowly with time. What you can see as plotted are the peripheries traced out descending with time .

But first quickly watch some fireworks to see what I am discussing about:

It is common experience to see fire ball bright splinters expanding to bigger circles as the entire cluster comes down slowly with time. The periphery is a portion of a circle whose radius increases. Center of circle is always descending by gravity.

The two parameters are $\theta, t$ angle of elevation at first burst or fire, and time $t$.

By C-discriminant method the parabola envelope is the eliminant of either $\theta$ variable or

$$F(x,y,\theta) =0 ,\, F_{\theta } (x,y,\theta) =0 \tag{6}$$

or $t$ time variable.

$$F(x,y, t) =0 ,\, F_{t } (x,y, t) =0 \tag{7}$$

The first one is already discussed, the second one is expanding/descending fireworks circles as already stated.

In the latter case working is:

$$x = v t \cos \theta , y = v t \sin \theta - g t^2/2 \tag{8}$$

$$(\frac{x}{vt})^2 + (\frac{y+ gt^2/2}{vt}) ^2 = 1 \tag{9}$$

$$x^2 + ( y + g t^2/2)^2 = v^2t^2 \tag{10}$$

which is a Circle.

To find its envelope, as before partially differentiate with respect to time $t$ and cancel $2t$ on either side of equation , bring $v^2/g$ to right side :

$$y + gt^2/2 = v^2/g \tag {11}$$

$$x^2 + (v^2/g)^2 = 2 v^2/g * ( v^2/g-y) \tag {12}$$

$$x^2 + ( 2 H)^2 += 4 H ( v^2/g -y )\tag{13}$$

$$x^2 = 4 H ( H-y) \tag{14}$$

which is the same parabola envelope obtained earlier with $\theta$ as parameter. End points $( x=0, y=H ; x= 2 H, y= 0 )$

Although circle end traces are visible in a fireworks display it needs imagination as before with variable gun barrel angle to see that each circle is tangent to a fixed envelope. Hope you enjoyed it.

• Thanks for your solution and insight on the parabolic mirror! (+1). Nice graphics too! In addition to different parabolas, why do you also plot partial concentric circles? – hypergeometric Oct 25 '15 at 5:56
• Also, is there an argument which would directly conclude that the envelope is a parabola with focus at launch point and tip at the peak of the vertical launch, without first going through the analysis? – hypergeometric Oct 25 '15 at 6:03
• The question in my comment above has been answered here. – hypergeometric Sep 30 '18 at 17:14


GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 27 Jan 2020, 20:38

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Each of 60 cars is parked in one of three empty parking lots

Author Message
TAGS:

### Hide Tags

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15994
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

30 Sep 2019, 20:27
00:00

Difficulty:

15% (low)

Question Stats:

89% (01:53) correct 11% (01:27) wrong based on 62 sessions

### HideShow timer Statistics

EMPOWERgmat PS Series:
Block 1, Question 5

Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?

A. 12
B. 20
C. 22
D. 28
E. 30

48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!) Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question SC Block 1 question pack will win an$85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.

To be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct

The OA and official explanation will be held until the 48 hour window has lapsed.

_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
VP
Joined: 20 Jul 2017
Posts: 1266
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

30 Sep 2019, 20:43
1
2
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5

Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?

A. 12
B. 20
C. 22
D. 28
E. 30

48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!) Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question SC Block 1 question pack will win an$85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.

To be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct

The OA and official explanation will be held until the 48 hour window has lapsed.

Let the number of cars parked in largest lot = x
--> Middle lot = x - 8 & Smallest lot = x - 16
So, x + (x - 8) + (x - 16) = 60
--> 3x - 24 = 60
--> 3x = 84
--> x = 28

IMO Option D
Senior Manager
Joined: 28 Feb 2014
Posts: 280
Location: India
GMAT 1: 570 Q49 V20
GPA: 3.97
WE: Engineering (Education)
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

30 Sep 2019, 21:25
1
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5

Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?

A. 12
B. 20
C. 22
D. 28
E. 30

let a,b,c be lot size in ascending order
a+b+c=60
c=b+8
c=a+16
Substituting
c-16+c-8+c=60
c = 28

D is correct
Intern
Joined: 09 Jul 2018
Posts: 12
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

02 Oct 2019, 16:42
1
Let the number of cars parked in largest lot = L
Middle lot = M & Smallest lot = S
But M=L-8 & S=L-16
So, L + (L - 8) + (L - 16) = 60
3L - 24 = 60
3L = 84
L = 28

IMO Option D
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15994
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

03 Oct 2019, 11:34
OFFICIAL EXPLANATION

Hi All,

We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...

Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...

IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4234
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5

Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?

A. 12
B. 20
C. 22
D. 28
E. 30

Let x = number of cars in the LARGEST lot

Given: the largest lot holds 8 more cars than the middle lot.
So, x - 8 = number of cars in the MIDDLE lot

Given: the largest lot holds 16 more cars than the smallest lot.
So, x - 16 = number of cars in the SMALLEST lot

Given: Each of 60 cars is parked in one of three empty parking lots.
We can write: (# of cars in LARGEST lot) + (# of cars in MIDDLE lot) + (# of cars in SMALLEST lot) = 60
Substitute values: (x) + (x - 8) + (x - 16) = 60
Simplify: 3x - 24 = 60
Add 24 to both sides: 3x = 84
Divide both sides by 3: x = 28

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Senior Manager
Joined: 05 Jul 2018
Posts: 459
Location: India
Concentration: General Management, Technology
GMAT 1: 600 Q47 V26
GRE 1: Q162 V149
GPA: 3.6
WE: Information Technology (Consulting)
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

10 Oct 2019, 22:39
1
The 3 slots A + B + C total to 60 cars

The highest slot C has 8 more than B and 16 more than A

Let us look at the second-largest answer choice which says C=28

As per question A=12 (16 less than C), B=20(8 less than C) ...This gives a total of 60. Hence Correct answer is D
EMPOWERgmatRichC: This is the correct way to do TTA ...right
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15994
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

11 Oct 2019, 11:30
Hi saukrit,

YES - that's a perfect use of TEST THE ANSWERS. In these sorts of situations, that Tactic (and some basic Arithmetic) is almost always faster and easier that a typical Algebraic approach - so it's something to keep in mind (especially if you have a pacing problem in the Quant section).

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4868
Location: India
GPA: 3.5
Re: Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

12 Oct 2019, 00:09
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5

Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?

A. 12
B. 20
C. 22
D. 28
E. 30

48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!) Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question SC Block 1 question pack will win an$85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.

To be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct

The OA and official explanation will be held until the 48 hour window has lapsed.

$$L = M + 8 = S + 16$$

Again , L + M + S = 60

Now, Plug in the options and check

(A) Can be straightaway rejected as value of L must be > 16 , Since in this case if L = 12 = S + 16 , Or, S = - 4 ( Not Possible )
(B) If L = 20 , M = 12 & S = 4 , L + M + S = 36
(C) If L = 22 , M = 14 & S = 6 , L + M + S = 42
(D) If L = 28 , M = 20 & S = 12 , L + M + S = 60
(E) If L = 30 , M = 22 & S = 14 , L + M + S = 66

_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Intern
Joined: 06 Sep 2019
Posts: 2
Each of 60 cars is parked in one of three empty parking lots  [#permalink]

### Show Tags

12 Oct 2019, 01:42
As the middle lot has 8 cars less than the larger it would be X-8
Smaller Lot is having 16 cars less than the larger that means X-16
As all the lots together having 60 it would be X+(X-8)+(X-16)=60
On solving this it becomes X= 28 Cars
Hence Option D would be apt.
Each of 60 cars is parked in one of three empty parking lots   [#permalink] 12 Oct 2019, 01:42
Display posts from previous: Sort by


# Fibonacci sequence: Given $n$ and $\mathrm{Fib}(n)$, is it possible to calculate $\mathrm{Fib}(n-1)$?

Given $n$ and $\newcommand{\Fib}{\mathrm{Fib}} \Fib(n)$, is it possible to calculate the previous number in the Fibonacci sequence - $\Fib(n-1)$ using integer math in constant time? In other words, I believe I'm looking for a closed form formula.

For example: Given $n=10$ and $\Fib(10)=55$, I'd like to determine that $\Fib(9)=34$ without using $\Fib(7) + \Fib(8)$.

• Even output of $F_{n-1}$ cannot be done in constant time. (The length of binary representation of $F_{n-1}$ is linear in $n$.) But based on the wording in original post I would guess that you mean using constantly many operations with integers. Aug 20 '14 at 8:14
• @Martin Sleziak - you are correct that I meant "constantly many operations with integers" Aug 20 '14 at 13:23

By Binet's Formula, we know that $$F_n = \dfrac{1}{\sqrt{5}}\left(\phi^n-(-\phi)^{-n}\right) \approx \dfrac{1}{\sqrt{5}}\phi^n$$ where $$\phi = \dfrac{1+\sqrt{5}}{2}$$.

Using this formula, we can get that $$F_{n-1}$$ is the nearest integer to $$\dfrac{1}{\phi} F_n$$ for large $$n$$

More rigorously, we have $$F_{n-1} - \dfrac{1}{\phi} F_n = \dfrac{1}{\sqrt{5}}\left(\phi^{n-1}-(-\phi)^{-(n-1)}\right) - \dfrac{1}{\phi\sqrt{5}}\left(\phi^n-(-\phi)^{-n}\right)$$ $$= -\dfrac{(-\phi)^{-(n-1)}}{\sqrt{5}} - \dfrac{(-\phi)^{-(n+1)}}{\sqrt{5}}$$ $$= \dfrac{(-\phi)^{-n}}{\sqrt{5}}\left(\phi + \phi^{-1}\right) = (-\phi)^{-n}$$.

Thus, for $$n \ge 2$$, we have $$\left|F_{n-1} - \dfrac{1}{\phi} F_n \right| = \phi^{-n} < \frac{1}{2}$$, and so, $$F_{n-1}$$ is the nearest integer to $$\dfrac{1}{\phi}F_n$$.

Dividing $$F_n$$ by $$\phi$$ and rounding the result to the nearest integer shouldn't be too computational.

• Here is a table that shows the pattern:$$\begin{array} {c|c|c} n & a_n & a_n / a_{n-1} \\ \hline 0 & 0 & \\ \hline 1 & 1 & \\ \hline 2 & 1 & 1.0000 \\ \hline 3 & 2 & 2.0000 \\ \hline 4 & 3 & 1.5000 \\ \hline 5 & 5 & 1.6667 \\ \hline 6 & 8 & 1.6000 \\ \hline 7 & 13 & 1.6250 \\ \hline 8 & 21 & 1.6154 \\ \hline 9 & 34 & 1.6190 \\ \hline 10 & 55 & 1.6176 \\ \hline 11 & 89 & 1.6182 \\ \hline 12 & 144 & 1.6180 \\ \hline 13 & 233 & 1.6181 \\ \hline 14 & 377 & 1.6180 \\ \hline \end{array}$$ Aug 20 '14 at 2:25
• @DanielV - agreed, this will work but as marty cohen points out in his anti-answer below, for large values of n simply rounding to the nearest integer will not necessarily yield the correct answer. Aug 20 '14 at 13:34

This is a dissenting note on some of these answers. I am purposly making this an answer, though it really is an anti-answer.

The problem with any answer that uses $\phi$ is that, as $n$ gets larger, $\phi$ has to be computed with increasingly great precision. This will not take constant time, either for the computation of $\phi$ or the multiplication by $\phi$.

Hint $\rm\ n\:$ is a Fibonacci number iff the interval $\rm\ [\phi\ n - 1/n,\ \phi\ n + 1/n]\$ contains a positive integer (the next Fibonacci number for $\rm\ n > 1)$. This follows from basic properties of continued fractions.

For example, $\rm\ 2 \to [2.7,\ 3.7],\ \ 3\to [4.5,\ 5.2],\ \ 5 \to [7.9,\ 8.3]\ \ldots$

For a proof see e.g.  T. Komatsu: The interval associated with a fibonacci number. $\$


# Thread: Geometric series problem .

1. ## Geometric series problem .

A house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?

My attempt :

It forms a sequence as follows :

$\displaystyle 50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$

The 15th term , $\displaystyle T_15=50000(1.09)^14=167086.35$
which is the amount he has to pay by the end of 15 years .

Thus , every month he has to pay $\displaystyle \frac{167086.35}{15\times12}=928.26$

My answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..

2. ## Repayments

Originally Posted by mathaddict
A house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?

My attempt :

It forms a sequence as follows :

$\displaystyle 50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$

The 15th term , $\displaystyle T_15=50000(1.09)^14=167086.35$
which is the amount he has to pay by the end of 15 years .

Thus , every month he has to pay $\displaystyle \frac{167086.35}{15\times12}=928.26$

My answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..
Thanks for showing us your working here. What you're forgetting is that
each year he pays not only the interest owing that year, but he also pays back some of the amount borrowed.

Are you allowed to use a spreadsheet to work out this answer? It's the easiest way.

If not, you'll have to let the monthly payment be x, and then say that, at the start of each new year, the amount owing

= amount owed at the start of the last year + interest payable during that year - 12x

Then work out how much interest he'll have to pay in the coming year to cover this new amount owing. Then repeat for the following year, and so on, up to the start of year 16. Then say that at the start of year 16 he owes nothing.

This is an Amortization problem and requires a special formula.

I've seen this type of problem posted many times before.
It seem that the students are never given the formula,
. . nor are they prepared for the (very) long derivation.

Could it be that all those teachers are ignorant of the difficulties
. . involved with time-payment problems?
I'm beginning to suspect that this is so . . .

A house buyer borrows $50,000 from a bank to buy a house. The rate of interest 9% per annum. The buyer is required to repay his loan in monthly installments for 15 years. Assuming that the rate of interest is fixed for the entire duration of the loan, find the amount per month he has to repay the bank? The formula is: .$\displaystyle A \;=\;P\,\frac{i(1+i)^n}{(1+i)^n-1}$. . where: .$\displaystyle \begin{Bmatrix}P &=& \text{principal} \\ i &=& \text{periodic interst rate} \\ n &=& \text{number of periods} \\ A &=& \text{periodic payment} \end{Bmatrix}$We have: .$\displaystyle \begin{array}{ccc}P \:=\:50,\!000 \\
i \:=\:\frac{9\%}{12} \:=\:0.0075 \\ n \:=\:15\!\cdot\!12 \:=\:180 \end{array}$Hence: .$\displaystyle A \;=\;50,\!000\,\frac{0.0075(1.0075)^{180}}{1.0075^ {180}-1} \;=\;507.1332921$Therefore, the monthly payment is: .$\displaystyle \boxed{\$507.13}$

4. ## Re :

Very big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?

5. Originally Posted by mathaddict
Very big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?
Check this and especially this out. It is one way of deriving it. There are or course other ways.
Go through page 70 up to page 99.
Or you can just order this book if get tired of reading it online.

6. ## Repayments

In fact, the interest is calculated at the start of each year for the whole year, as if the whole amount is owing for the whole year. So you have to use the values

P = 50000
i = 0.09
n = 15

to get the total payment each year, and then divide by 12, to get a monthly payment of 516.91.

Incidentally, I don't think it's really that hard to work out the formula for yourself. After
$\displaystyle n$ years you'll find that the amount owing is

$\displaystyle P(1+i)^n - [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x$, where $\displaystyle x$ is the annual payment

Inside the square brackets, it's a GP whose sum is
$\displaystyle \frac{(1+i)^n - 1}{i}$

Set this sum equal to zero and solve for
$\displaystyle x$ to obtain the formula.

7. Thanks all but i am having some problems with the series .

[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math

$\displaystyle T_1=x$

$\displaystyle T_2=xi+x=x(1+i)$

$\displaystyle T_3=[x(1+i)]\cdot{i}+x$ but this is not equal to $\displaystyle x(1+i)^2$

Why is it so ??

8. ## GP

Originally Posted by mathaddict
Thanks all but i am having some problems with the series .

[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math

$\displaystyle T_1=x$

$\displaystyle T_2=xi+x=x(1+i)$

$\displaystyle T_3=[x(1+i)]\cdot{i}+x$ but this is not equal to $\displaystyle x(1+i)^2$

Why is it so ??
Set your working out like this:

At the end of year 1, the amount owed is the original sum borrowed + the interest for the year - the total payments during the year

$\displaystyle = P + Pi - x$

$\displaystyle = P(1 + i) - x$

This, then is the amount owing at the start of year 2: the 'new $\displaystyle P$' if you like. So at the end of year 2, we'll replace $\displaystyle P$ by $\displaystyle P(1+i) - x$, to get that the amount owing is:

$\displaystyle [P(1 +i) - x](1+i) - x$

$\displaystyle = P(1+i)^2 - x[1 + (1 + i)]$

Continue like this, multiplying by $\displaystyle (1+i)$ and subtracting $\displaystyle x$ each time. So at the end of the year 3, the amount owing is:

$\displaystyle P(1+i)^3 - x[1 + (1 + i) + (1 +i)^2]$

And, in a similar way, at the end of year $\displaystyle n$, it is:

$\displaystyle P(1+i)^n - x[1 + (1 + i) + ...+ (1 +i)^{n-1}]$

So, inside the square brackets, [...], we have a GP whose first term is $\displaystyle 1$, common ratio $\displaystyle (1+i)$ and number of terms = $\displaystyle n$, and whose sum is $\displaystyle \frac{(1+i)^n - 1}{(1+i) - 1}= \frac{(1+i)^n - 1}{i}$.

Now if the whole sum is paid of after $\displaystyle n$ years, this total is zero. I.e.

$\displaystyle P(1+i)^n - x\frac{(1+i)^n - 1}{i} = 0$

Hence $\displaystyle x = \frac{iP(1+i)^n}{(1+i)^n - 1}$

OK now?


# Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X

Let $$X$$ be a random variable with a continuous and strictly increasing c.d.f. $$F$$ (so that the quantile function $$F^{−1}$$ is well-deﬁned). Deﬁne a new random variable $$Y$$ by $$Y = F(X)$$. Show that $$Y$$ follows a uniform distribution on the interval $$[0, 1]$$.

My initial thought is that $$Y$$ is distributed on the interval $$[0,1]$$ because this is the range of $$F$$. But how do you show that it is uniform?

• This is not true in cases where there's a discrete component. For example, suppose $X=\left.\begin{cases} 1/2 & \text{with probability }1/2, \\ W & \text{with probability }1/2,\end{cases}\right\}$ and $W$ is uniformly distributed on $[0,1]$, and that the choice between whether $X=1/2$ or not is independent of $W$. Then the cdf of $X$ has no values between $1/4$ and $3/4$, so it cannot be uniformly distributed on $[0,1]$. It is, however, true of continuous distributions. Jul 15, 2014 at 21:41
• see the text of the question. X is continuous! Jul 15, 2014 at 21:44
• By the way, it is not necessary that $F$ is a strictly increasing CDF, continuity is sufficient. Just define the quantile function the usual way as a generalized inverse via $F^-(y)=inf\{x\in\mathbb{R}: F(x)\geq y\}$. See the proof of Proposition 3.1 in Embrechts, P., Hofert, M.: A note on generalized inverses. Mathematical Methods of Operations Research 77(3), 423-432 link for a very careful and detailed explanation. Jul 16, 2014 at 14:38
• Thanks @binkyhorse - that reference is really good. Apr 5, 2015 at 22:33
• @s0ulr3aper07 By Proposition 3.1 in the paper I linked above, yes. Prop. 3.1: Let F be a distribution function and X ~ F. (a) If F is continuous, F(X)∼U[0,1]. The paper includes a detailed proof. Mar 27, 2019 at 20:10

Let $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \in [0,1]$ we have:

$F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$.

What distribution has this CDF?

• Are all CDF's of continuous densities invertible? Aug 22, 2017 at 7:43
• @tintinthong: not always completely, but enough. If you define $G(y)= \inf\{x:F_X(x) \ge y\}$ then $F_X(G(y))=y$ when $y \in (0,1)$ Aug 22, 2017 at 22:12
• Strictly increasing and continuous CDF is needed. Feb 20, 2019 at 15:36

$$Prob(Y\leq x)=P(F(X)\leq x)=P(X\leq F^{-1}(x))=x \\$$ The last equality is from the definition of the quantile function.

Let $y=g(x)$ be a mapping of the random variable $x$ distributed according to $f(x)$. In the mapping $y=g(x)$ you preserve the condition of probability density (namely you counts the same number of events in the respective bins)

$$h(y)dy=f(x)dx$$

where h(y) is the probability distribution of $y$

if $h(y)=1$ (uniform distribution) we have

$$dy=g'(x)dx=f(x)dx$$

This means that $$g(x)=\int f(x)dx$$

namely the function $g(x)$ that maps the random variable $x$ distributed according $f(x)$, into a random variable $y$ distributed uniformly is his own cumulative distribution function $\int f(x) dx$.

Here is an approach that does not use the quantile function whatsoever - the only property used is that independent copies of $$X$$ have zero probability of being equal. (The main ingredient in my argument is conditional expectation.)

Consider the cumulative distribution function of $$X$$, namely $$F(t)=\mathbb P(X\leq t).$$ Your random variable - which I will suggestively call $$U$$ instead of $$Y$$ - can be described by starting with two independent and identically distributed random variables $$X,Z$$ and considering the conditional probability $$U=\mathbb P(X\leq Z\mid Z).$$ Then, for all integers $$n\geq 1$$, we can represent $$U^n$$ as follows. Let $$X_1,X_2,\ldots,X_n,Z$$ be independent and identically distributed. By independence, $$\mathbb P\bigl(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z\bigm\vert Z\bigr)=U^n,$$ and thus by the tower property $$\mathbb EU^n=\mathbb P(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z)=\mathbb P\bigl(Z=\max(X_1,X_2,\ldots,X_n,Z)\bigr).$$ Since $$X_1,\ldots,X_n,Z$$ are iid, each of them is equally likely to be the maximum and therefore $$\mathbb EU^n=\frac{1}{n+1}.$$ Thus $$U$$ has the same moments as a uniformly distributed random variable on $$[0,1]$$. Since $$U$$ is supported in $$[0,1]$$ as well, it follows (by the uniqueness of the Hausdorff moment problem) that $$U$$ is uniformly distributed, as desired.

Let $$y\in(0,1)$$. Since $$F$$ is continuous, there exists $$x\in\mathbb{R}$$ s.t. $$F(x)=y$$. Thus, $$\mathsf{P}(Y\le y)=\mathsf{P}(F(X)\le F(x))=F(x)=y,$$ i.e., $$Y\sim\text{U}[0,1]$$. In order to see the first equality we don't need continuity. Specifically, since any cdf is right-continuous, \begin{align} \{F(X)\le F(x)\}&=\{\{F(X)\le F(x)\}\cap\{X\le x\}\}\cup \{\{F(X)\le F(x)\}\cap\{X>x\}\} \\ &=\{X\le x\}\cup \{\{F(X)=F(x)\}\cap\{X>x\}\}, \end{align} and $$\mathsf{P}(\{F(X)=F(x)\}\cap\{X>x\})=0$$.

For a proof of this problem when $$F_X(x)$$ is strictly increasing, refer to JimmyK4542's answer. Let's assume $$F_X(x)$$ is just non-decreasing (there are intervals such as $$[a,b]$$ where $$F_X(x') = c$$ for $$x'\in[a,b]$$). We define $$G(y)$$ similar to what Henry's comment suggests: $$G(y)=\inf\{x:F_X(x)\gt y\}$$ Now substituting this expression in what Jimmy has written will give us: $$F_Y(y) = \Pr[Y \le y] = \Pr[F_X(X) \le y] = \Pr[X \le G(y)] = F_X(G(y))= y \label{eq:I}\tag{I}$$

We need to show that:

1. $$F_X(x)\le y \rightarrow x \le G(y)$$
2. $$F_X(G(y))=y$$

The second argument is easier to prove. We have the following expression almost according to definitions: $$F_X(G(y))= F_X(\inf\{x:F_X(x)\gt y\})= y$$ Now for the first argument, we can still use what $$G(y)=\inf\{\cdots\}$$ implies; if $$F_X(x)\le y$$, then $$x\le \inf\{x:F_X(x)\gt y\};$$ hence $$x\le G(y)$$.

With the two arguments proved and a substitution in \ref{eq:I}, we have proved the main argument.

• I personally believe this problem is a severe case for abuse of notation, and a bad professor's problem. Feb 19 at 1:46


+0

# what is the probability of winning a lottery which requires selection of 5 numbers from1 to 40

0
895
7

what is the probability of winning a lottery which requires selection of 5 numbers from1 to 40

May 7, 2014

#2
+17
+8
Calculate the numerator: There is only one winning combination. So the numerator of the desired probability is 1. Calculate the denominator: There are "40 numbers, choose 5", or 40C5 possible combinations. So the denominator of the desired probability is 40C5. So the probability is 1/40C5=1/658008

May 7, 2014

#1
+1006
0

For this kind of probability problem, assuming the number is removed from play once chosen, the probability increases by one each time.

$$\left({\frac{{\mathtt{1}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right)$$

The bottom five numbers multiply to 78960960, which means that the odds are 1/78960960

May 7, 2014
#2
+17
+8
Calculate the numerator: There is only one winning combination. So the numerator of the desired probability is 1. Calculate the denominator: There are "40 numbers, choose 5", or 40C5 possible combinations. So the denominator of the desired probability is 40C5. So the probability is 1/40C5=1/658008

FoxTears May 7, 2014
#3
+17
0

Mine was assuming the numbers can be reused

May 7, 2014
#4
+115418
+5

The number of 5 digit combinations (order doesn't matter) from 40 is

40C5

$${\left({\frac{{\mathtt{40}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{40}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{658\,008}}$$

only one of those wins so the chance is     $$\frac{1}{658008}$$

Just like FoxTears said!

Thanks Foxtears.

Foxtears, I don't think that yours was assuming that the numbers could be reused.

It was just assuming that the order they were chosen in was of no consequence.

GoldenLeaf I believe yours would have been correct if the numbers had to be chosen in a specific order.

May 7, 2014
#5
+32772
+5

GoldenLeaf's approach was almost right, but he should have had 5, 4, 3, 2 and 1 in the numerators.

There are 5 possible choices for the first number, so probability of this is 5/40.  For each of these

there are 4 possible choices for the 2nd number, so probability of both 1st and 2nd is (5/40)*(4/39).  For each of these ... etc. to get the overall probability as

$${\mathtt{p}} = \left({\frac{{\mathtt{5}}}{{\mathtt{40}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{39}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{38}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{37}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{36}}}}\right) \Rightarrow {\mathtt{p}} = {\mathtt{0.000\: \!001\: \!519\: \!738\: \!361\: \!8}}$$

or p≈1.52*10-6

1/p = 658008

May 7, 2014
#6
+115418
0

Yes of course - that did not occur to me.  Thanks Alan.

May 7, 2014
#7
0

Calling Bertie, CPhill and Rom (Troll alert! General quarters!)

Along with Melody and Alan, we need input from Bertie, CPhill, and Rom, too.

With the top brains represented here that Troll might comeback and bite one or more.

I know he’ll bite FoxTears for saying, “Mine was assuming the numbers can be reused” (your formula (40C5) contradicts this).

The Troll will follow-up, by asking what is the probability of (something) of (five, six or seven) mathematicians getting a probability question correct.

Then he’ll say about the same as walking, blindfolded, through a large buffalo heard and stepping in four piles with both feet.

What’s annoying: he’s usually right.

CPhill might not be able to make it. He might be trying to get the “Roman Zero” from inside Sisyphus’s boulder.

Maybe he could take a break and visit. He’s not been around for a day or two. I hope he wasn’t squashed by the boulder.

No matter: cartoon physics seem to apply in that realm. If he does visit, maybe one of those angels will help him get the “Roman Zero.”

By: Troll Detector

May 7, 2014