MATH 201 Recitation Problems

Solution

(a) Since $Y \sim \text{Geom}(\frac{1}{6})$, ${\rm{E}}[Y]=6$, moreover geometric random variables take nonnegative values, hence we can apply Markov’s inequality to get

(b) Since $Y \sim \text{Geom}(\frac{1}{6})$, ${\rm{Var}}(Y)=\frac{1-1/6}{(1/6)^2}=30$, and applying Chebyshev’s inequality, we get

(c) Since $Y \sim \text{Geom}(\frac{1}{6})$,

Solution (a) Since ${\rm{E}}[X]=10$, moreover $X$ take nonnegative values, hence we can apply Markov’s inequality to get \(\begin{align*} {\rm{P}}(X\geq 15) \leq \frac{{\rm{E}}(X)}{15}=\frac{10}{15} \approx 0.666. \end{align*}\)

(b) Since ${\rm{Var}}(X)=3$, and applying Chebyshev’s inequality, we get \(\begin{align*} {\rm{P}}(X\geq 15) ={\rm{P}}(X-10\geq 5) \leq {\rm{P}}(|X-10|\geq 5) \leq \frac{{\rm{Var}}(X)}{5^2}=\frac{3}{25}=0.12. \end{align*}\)

(c) Let $S_n=\sum_{k=1}^n Y_i$. Since $Y_i$’s are iid with finite mean and variance, we can apply the CLT to get \(\begin{align*} \rm{P}(S_{300}\geq 3030) &=\rm{P}\left(\frac{S_{300}-(300)(10)}{\sqrt{(300)(3)}}\geq \frac{3030-(300)(10)}{\sqrt{(300)(3)}} \right)\\&=\rm{P}\left(\frac{S_{300}-3000}{30}\geq 1 \right)\approx 1-\Phi(1)=1 - 0.8413=0.1587 \end{align*}\)

Solution

Let $X_i$ be waiting time for the bus 5 on the $i$-th day and $Y_i$ waiting time for the bus 8 on the $i$-th day. Then we are given that $X_i \sim \text{Exp}(\frac{1}{10})$ and $Y_i \sim \text{Exp}(\frac{1}{20})$. Now let us define the random variable \(\begin{align*} J_i:=\min(X_i,Y_i)\hskip 5pt \text{and }\hskip 5pt I_i:=I_{(X_i<Y_i)}. \end{align*}\)

Then we have $S_n=\sum_{i=1}^nJ_i$ and $T_n=\sum_{i=1}^n I_i$. Recall from Chapter 6 (exercises 6.33 and 6.34), we know $J_i \sim\text{Exp}(\frac{1}{10}+\frac{1}{20})=\text{Exp}(\frac{3}{20})$ and $I\sim \text{Ber}(p)$ where $p={\rm{P}}(X_i<Y_i)=\frac{1/10}{3/20}=\frac{2}{3}. $

(a) From Law of large numbers we know for any $\epsilon>0$

and \(\begin{align*} { \rm{P}}\left(\left|\frac{S_n}{n}-\mu\right|\leq \epsilon\right) \to 1 \text{ as }n\to \infty. \end{align*}\)

Now, take $\epsilon=7-\frac{20}{3}=\frac{1}{3}$, then

Hence

(b) Take $\epsilon=-0.6+\frac{2}{3}=\frac{1}{15}$, then

Hence

Solution

(a) Note that $M’_X(t)=-\frac{4}{3}e^{-4t} + \frac{5}{6} e^{5t}$ and $M’‘_X(t)=\frac{16}{3}e^{-4t} + \frac{25}{6} e^{5t}$ and hence

(b) The possible values of $X$ are ${-4,0,5}$ and

\(\begin{align*} {\rm{P}}(X=-4)=\frac{1}{3}, \hskip 5pt {\rm{P}}(X=0)=\frac{1}{2}, \hskip 5pt {\rm{P}}(X=5)=\frac{1}{6}. \end{align*}\) Hence

Solution

(a)

(b) Note that for $t<1$,

and in general for $t<1$

\(\begin{align*} M^{(n)}_X(t)=(n+1)!(1-t)^{-(n+2)}. \end{align*}\) So, we see

\(\begin{align*} {\rm{E}}[X]=M'_X(0)=2(1)^{-3}=2, \hskip 5pt {\rm{E}}[X^2]=M''_X(0)=(3)(2)=6, \hskip 5pt {\rm{E}}[X^3]=M'''_X(0)=(4)(3)(2)=24 \end{align*}\) and in general for $n\in {\mathbb{N}}$

Solution

Define for $i\in {1,2,\dots,29,30}$

\(\begin{align*} I_i=\begin{cases} 1, &\text{if ith flip is heads},\\ 0, &\text{if ith flip is tails}. \end{cases} \end{align*}\) Then $X=I_1+\dots+I_{20}$ and $Y=I_{11}+\dots+I_{30}$ where $I_i \sim \text{Ber}(\frac{1}{2})$ and all flips are independent. So we have ${\rm{E}}[I_i]={\rm{E}}[I_1]=\frac{1}{2}$, ${\rm{E}}[I_i^2]={\rm{E}}[I_1^2]=\frac{1}{2}$, ${\rm{Var}}(I_i)={\rm{Var}}(I_1)=\frac{1}{4}$, and for any $i\neq j $

Using the linearity of the covariance in each component, we see

Using the independence we have

\(\begin{align*} {\rm{Var}}(X)={\rm{Var}}(I_1+\dots+I_{20})={\rm{Var}}(I_1)\dots{\rm{Var}}(I_{20})=\frac{20}{4}=5. \end{align*}\) Similarly,

\(\begin{align*} {\rm{Var}}(Y)={\rm{Var}}(I_{11}+\dots+I_{30})={\rm{Var}}(I_{11})\dots{\rm{Var}}(I_{30})=\frac{20}{4}=5. \end{align*}\) Finally,

Solution Recall the probability mass function of the Poisson distribution:

\(\begin{align*} {\rm{P}}(X=k)=p(k)=e^{-\lambda} \frac{\lambda^k}{k!} \text{ for }k=0,1,2,\dots. \end{align*}\) Then

To compute this sum we used the Taylor series of exponential function, and integrate both sides to get

Solution

For the number of times an event occur on a given interval, it is appropriate use Poisson distribution with mean $\lambda=6$ as the probability of an error is relatively low. Then

Solution

Before we do any calculation first we determine the region where $f(x,y)$ is non-zero. If we plot the lines $y=x$ and $y=2-x$, we see that they intersect at $(1,1)$. The constrains $0<y<1$ and $y<x<2-y$ tell us that the region is the triangle with vertices $(0,0), (1,1), (2,0)$.

First we find $f_Y(y)$ by integrating against $x$. For a fixed value of $y \in [0,1]$, the region is bounded by $x=y$ and $x=2-y$, so

so that

To find the marginal of $X$ we integrate $f$ against $y$. There are three cases.

Case (1): when $0\leq x \leq 1$, we integrate

Case (2): when $1 < x \leq 2$, we integrate

The last case corresponds to $f_X(x) = 0 $ for $x$ outside of $[0,2]$, so that

Solution

(a) Probability mass function of $X$ can be obtained by summing the rows: \(\begin{align*} {\rm{P}}(X=1)&=\sum_{l=0}^3p_{(X,Y)}(1,l)=p_{(X,Y)}(1,0)+p_{(X,Y)}(1,1)+p_{(X,Y)}(1,2)+p_{(X,Y)}(1,3)\\&=\frac{1}{15}+\frac{1}{15}+\frac{2}{15}+\frac{1}{15}=\frac{1}{3}\\ {\rm{P}}(X=2)&=\sum_{l=0}^3p_{(X,Y)}(2,l)=p_{(X,Y)}(2,0)+p_{(X,Y)}(2,1)+p_{(X,Y)}(2,2)+p_{(X,Y)}(2,3)\\&=\frac{1}{10}+\frac{1}{10}+\frac{1}{5}+\frac{1}{10}=\frac{1}{2}\\ {\rm{P}}(X=3)&=\sum_{l=0}^3p_{(X,Y)}(3,l)=p_{(X,Y)}(3,0)+p_{(X,Y)}(3,1)+p_{(X,Y)}(3,2)+p_{(X,Y)}(3,3)\\&=\frac{1}{30}+\frac{1}{30}+0+\frac{1}{10}=\frac{1}{6}. \end{align*}\)

Similarly,

Probability mass function of $Y$ can be obtained by summing the columns:

(b) The pairs $(l,k)$ satisfying $l+k^2\leq 2$ are $(1,0),(1,1),(2,0)$. So,

Solution

Multiple of 3s are $3,6$ which have probability $\frac{1}{3}$ Let $S\sim \text{Bin}(300,\frac{1}{300})$. Then ${\rm{E}}[S]=300\frac{1}{3}=100$ and ${\rm{Var}}[S]=300\frac{1}{3}\frac{2}{3}=\frac{200}{3}$. Using normal approximation with continuity correction, we see

Solution

Recall

In the set up of the question $n=81$, $\epsilon = 0.05$. Hence the confidence level is \(2\Phi(2(0.05) \sqrt{81})-1=2\Phi(0.9)-1\approx 2(0.8159)-1=0.6318.\)

Solution

Let $S_n$ denote the number of rounds you win and $X_n$ denote the total earnings. Then $S_n \sim \text{Bin}(n,\frac{1}{20})$ and $X_n=10S_n-(n-S_n)=11S_n-n$. First we would like to estimate ${\rm{P}}(X_{200}>-100)$. We will use normal approximation for $S_{200}$. Note that ${\rm{E}}[S_{200}]=200\frac{1}{20}=10$ and ${\rm{Var}}[S_{200}]=200\frac{1}{20}\frac{19}{20}=\frac{19}{2}$. Then, with continuity correction

The actual probability using a computer \({\rm{P}}(S_{200} \geq 10)=1-\sum_{i=0}^{9} \binom{200}{i}\left(\frac{1}{20}\right)^i\left(\frac{19}{20}\right)^{200-i} \approx 0.54529\)

Similarly, \({\rm{P}}(S_{300}\geq 10) \approx 0.1762\)

Solution

Note that for any $a,b$, \(\mathrm{E}[aX+b]=a \mathrm{E}[X]+b \text{ and }\mathrm{Var}(aX+b)=a^2 \mathrm{Var}(X)\)

(a) Hence

(b)

(c) Using $\mathrm{Var}(3X+4)= \mathrm{E}[(3X+4)^2]- \left(\mathrm{E}[3X+4]\right)^2$ and part a and b, we get

Solution

Let $X$ be the the number of the flips until the first head. Then we have $X\sim \text{Geo}(p)$. And let $Y=g(X)=3X+1$ which corresponds to the earnings of Erin. Then, we have

\(=3 p\sum_{k=1}^{\infty} k(1-p)^{k-1}+p\sum_{k=1}^{\infty} (1-p)^{k-1} = \frac{3p}{(1-(1-p))^2}+\frac{p}{1-(1-p)}=\frac{3}{p}+1\) where we have used the geometric sum

and its derivative

with $x=1-p$.

Solution

Note that $X\sim \mathcal{N}(3,4)$ and $Z=\frac{X-3}{2} \sim \mathcal{N}(0,1)$. Let $\Phi$ be the cdf of $Z$.

(a) \({\rm{P}}(2<X<6)= {\rm{P}}\left(\frac{2-3}{2}<\frac{X-3}{2}<\frac{6-3}{2}\right) ={\rm{P}}\left(-0.5<\frac{X-3}{2}<1.5\right)=\Phi(1.5)-\Phi(-0.5)\)

\(=\Phi(1.5)-(1-\Phi(0.5)) =0.9332-1+0.6915= 0.6247\) where we used $\Phi(-x)=1-\Phi(x)$ for $x>0$ and a table for $\Phi$.

(b) Note that

So we want to solve $1-\Phi(\frac{c-3}{2})=0.33$ or $\Phi(\frac{c-3}{2})=0.67$. From the table $\frac{c-3}{2}=0.44$ is a good approximation and hence $c=3.88.$

(c) \(\mathrm{E}[X^2]=\mathrm{Var}(X)+\left(\mathrm{E}(X)\right)^2=4+3^2=13.\)

Solution

(a) In order $f$ to be a pdf, we must have \(1= \int_{-\infty}^{\infty}f(x)dx=\int_1^2 \frac{1}{4}\,dx+\int_3^5 c\,dx=\frac{1}{4}+2c,\) Hence $2c=\frac{3}{4}$ and $c=\frac{3}{8}$.

(b) \({\rm{P}}(\frac{3}{2}<X<4)= \int_{\frac{3}{2}}^{4}f(x)dx=\int_{\frac{3}{2}}^2 \frac{1}{4}\,dx+\int_3^4 \frac{3}{8}\,dx=\frac{1}{8}+\frac{3}{8}=\frac{1}{2}.\)

(c) First, let us compute ${\rm{E}}[X]$: \({\rm{E}}[X]=\int_{-\infty}^{\infty} xf(x)\,dx= \int_{1}^2 \frac{x}{2}\,dx+\int_3^5 \frac{3x}{8}\,dx= \frac{3}{4}+\frac{3(16)}{16}=\frac{15}{4}.\)

Now, using ${\rm{E}[X]}$ and linearity of the expectation, we get \({\rm{E}}[8X+3]= 8{\rm{E}}[X]+3=8\frac{15}{4}+3=33.\)

Solution Let $X_1, X_2$ denote the draws so that $X=\max(X_1,X_2)$. Then for $k=2,3,4,5,6$ we have \(p(k)={\rm{P}}(X=k)={\rm{P}}(\max(X_1,X_2)=k)={\rm{P}}(\max(X_1,X_2)\leq k)-{\rm{P}}(\max(X_1,X_2)\leq k-1)\)

Hence, \({\rm{E}}[X]=\sum_{k=2}^6 k \frac{2k-3}{30} = \frac{1}{15} \sum_{k=2}^6 k^2 -\frac{1}{10} \sum_{k=2}^6 k =\frac{1}{15}(\frac{(13)(7)(6)}{6}-1) -\frac{1}{10}(\frac{(7)(6)}{2}-1)\)

Solution

(a) Let $Y=e^X$. We will find cdf of $Y$ first. For $1<y<e$ we have is \({\rm{P}}(Y\leq y)= {\rm{P}}(e^X\leq y)={\rm{P}}(X\leq \ln y) = \int_0^{\ln y} 1 dx=\ln y.\) By differentiating $F_Y$, we obtain the density function of $Y$ as \(f_Y(y)=\frac{1}{y}\, \hskip 2pt 1\leq y\leq e.\) Hence, \({\rm{E}}[Y]=\int_1^e y \frac{1}{y}\,dy=e-1.\) (b) Using the formula directly, we have \({\rm{E}}[e^X]=\int_{-\infty}^{\infty} e^x f(x)dx=\int_{0}^{1} e^x dx=e-1.\)

Solution

(a)

(b)

Solution

(a) Clearly $f$ is nonnegative and

\(\int_{-\infty}^{\infty}f(x)\,dx=\int_0^{\infty} e^{-x}\,dx=1.\) So, $f$ is a pdf.

(b) On one hand, \(\rmP(X>s+t\big| X>t)=\frac{ \rmP(X>s+t, X>t)}{\rmP(X>t)}=\frac{ \rmP(X>s+t)}{\rmP(X>t)}=\frac{\int_{s+t}^{\infty} e^{-x}\,dx }{\int_{t}^{\infty} e^{-x}\,dx}=\frac{e^{-(s+t)}}{e^{-t}}=e^{-s}\) and on the other hand \(\rmP(X>s)={\int_{s}^{\infty} e^{-x}\,dx}=e^{-s}.\)

Hence the identity follows.

Solution

Now for $s<1$, we have

For $1\leq s<2$, we have

For $2\leq s<3$, we have

For $3\leq s<4$, we have

Finally, for $s\geq 4$, we have

Collecting this information, we can write

Solution

Note that $#\Omega=6^2=36$and

(a) $A_1$ and $B$ are not independent since

(b) $A_2$ and $B$ are independent since

(c) $A_2, B,C$ are pairwise independent since

\(\rmP(A_2\cap B)=\frac{1}{36}= \frac{6}{36}\frac{6}{36}=\rmP(A_2)\rmP(B)\)
\(\rmP(A_2\cap C)=\frac{1}{36}=\frac{6}{36}\frac{6}{36}=\rmP(A_2)\rmP(C)\)
\(\rmP(B\cap C)=\frac{1}{36}= \frac{6}{36}\frac{6}{36}=\rmP(B)\rmP(C) .\)

(d) $A_2, B,C$ are \underline{not} independent although they are pairwise independent since

Solution

(a) It is easier to compute the probability of the complement, that is of no success in the first $n$ trials. If we let $A_i$ denote the event of a failure in the $i$th trial, then by independence we have

\(\rmP(A_1\cap A_2\dots A_n)=\rmP(A_1)\rmP(A_2)\dots \rmP(A_n)=(1-p)^n.\) Hence \(\rmP(\text{at least 1 success})=1-(1-p)^n.\)

(b) Any particular sequence of the n trials with $k$ successes and $(n-k)$ failures will occur with probability $p^k(1-p)^{n-k}$. As there are $\binom{n}{k}$ such sequences, we have

\(\rmP(\text{exactly }k \text{ successes})=\binom{n}{k} p^k(1-p)^{n-k}.\)

Solution

(a) If $A$ and $B$ are disjoint, then \(\rmP(A\cup B)=\rmP(A)+\rmP(B)=0.3+0.6=0.9.\)

(b) If $A$ and $B$ are independent, then \(\rmP(A\cup B)=\rmP(A)+\rmP(B)-\rmP(A\cap B)=\rmP(A)+\rmP(B)-\rmP(A)\rmP(B)=0.3+0.6-0.18=0.72.\)

Solution

Let \(B_k = \{k\text{th ball drawn is red}\}.\) Then \(\rm{P}(B)=\rm{P}\left(\bigcup_{k=1}^l B_k \right).\) Using inclusion/exclusion principle, we have \({\rm{P}}(B)={\rm{P}}\left(\bigcup_{k=1}^l B_k \right)=\sum_{k=1}^l(-1)^{k+1}\sum_{1\leq i_1<\dots <i_k \leq l} {\rm{P}}(B_{i_1}\cap \dots \cap B_{i_k}).\)Note that \({\rm{P}}(B_{i_1}\cap \dots \cap B_{i_k})=\frac{3^k}{n^k}.\)

Hence 

Note we can also compute this probability without inclusion/exclusion:

Solution

(a) $X$ can take values $1,2,3,4,5,6$.

(b) Note

(c) Using the hint $P(X = k) = P(X \leq k) - P(X \leq k-1)$, we have

\(P(X = k) = \frac{k^2}{36}-\frac{(k-1)^2}{36}=\frac{2k-1}{36}.\)

Solution

Let $A={\text{ Jim picks the unfair coin}}$ and $B={\text{he gets 5 tails in 5 tosses}}$. Note that \({\rm{P}}(A\cap B) =\frac{1}{100}\) and ${\rm{P}}(B)={\rm{P}}(\text{choosing a fair coin})\frac{1}{2^5}+{\rm{P}}(\text{choosing the unfair coin}) 1= \frac{1}{100}1+\frac{99}{100}\frac{1}{2^5}$. Then using the conditional probabilities, we see \({\rm{P}}(A\big|B)=\frac{{\rm{P}} (A\cap B)}{{\rm{P} (B)}}=\frac{\frac{1}{100}}{\frac{1}{100}1+\frac{99}{100}\frac{1}{2^5}}.\)

Solution

First number the balls from \(\{1,2,\dots,11\}\), where \(\{1,2,\dots,6\}\) corresponds to white balls and \(\{7,8,\dots,11\}\) to black balls. We pick 3 numbers and want to know the probability that 2 of them are from \(\{7,8,\dots,11\}\) and one of them is from \(\{1,2,\dots,6\}\).

Solution with order:

\(\Omega=\{(s_1,s_2,s_3):s_i\neq s_j \text{ for all }i\neq j\}\). Then \(\#\Omega=(11)_3=11\cdot 10\cdot 9\). Let \(A:=\{ (s_1,s_2,s_3) \in \Omega: \text{two of them} \in \{7,\dots, 11\} \text{ and } \text{ one of them } \in \{1,\dots, 6\}\}.\) Then \(\#A= 3 \cdot (5\cdot 4 \cdot 6)\) and hence \(\rmP(A)= \frac{3\cdot5\cdot 4\cdot 6}{11\cdot 10\cdot 9 } =\frac{4}{11}.\)

Solution without order:

Let \(\tilde{\Omega}=\{B\subset \{1,2,\dots,11\}: \#B=3\}\). Then \(\#\tilde{\Omega}=\binom{11}{3}\). Let \(\tilde{A}\in \tilde{\Omega}\) be such that

Then \(\#\tilde{A}=\binom{6}{1} \cdot \binom{5}{2}\) and hence

Solution

There are \(\binom{52}{13}\cdot \binom{39}{13}\cdot \binom{26}{13} \cdot \binom{13}{13}\) possible ways to divide the cards among 4 people. There are 4 ways to choose who gets all the spades and there are \(\binom{39}{13}\cdot \binom{26}{13} \cdot \binom{13}{13}\) possible ways to divide the rest of the cards among the rest 3 people. Hence

Solution

There are \(\binom{52}{13}\cdot \binom{39}{13}\cdot \binom{26}{13} \cdot \binom{13}{13}\) possible ways to divide the cards among 4 people. If we put aside the aces, there are \(\binom{48}{12}\cdot \binom{36}{12} \cdot \binom{24}{12} \cdot \binom{12}{12}\) possible ways to divide the rest of the cards among the rest 4 people. As there are \(4!\) ways of dividing the 4 aces, the desired probability is \(\rmP(A)=\frac{4!\cdot \binom{48}{12}\cdot \binom{36}{12} \cdot \binom{24}{12} \cdot \binom{12}{12} }{\binom{52}{13}\cdot \binom{39}{13}\cdot \binom{26}{13} \cdot \binom{13}{13}} \approx 0.1\)

Solution

(a) Using inclusion/exclusion principle, we have \(\rmP(E\cap F)=\rmP(E)+\rmP(F)-\rmP(E\cup F)= 0.9+0.8 - \rmP(E\cup F)\geq 1.7-1=0.7.\)

(b) Similarly using inclusion/exclusion principle, we have \(\rmP(E\cap F)=\rmP(E)\rmP(F)-\rmP(E\cup F)\geq \rmP(E)\rmP(F) - 1.\)

Solution

would represent 2 oranges in first bin, 2 oranges in the second bin and 1 orange in each of the third, fourth and fifth bin. Altogether, there are $7+5-1$ symbols. We have

\(=\binom{7+5-1}{7}=\binom{11}{7}\) many ways to put 7 oranges into 11 spots.

Solution (a)

(b) We can choose \(\binom{13}{5}\) ways the ranks of each card and choose \(\binom{4}{1}\) ways the suits of each card. Hence

(c) We can choose \(\binom{13}{2}\) ways the ranks of the pairs and choose \(\binom{4}{2}\) ways the suits of each pair and we can choose the last card in \(\binom{44}{1}\) ways. Hence

\(\binom{13}{2} \binom{4}{2}^2 \binom{44}{1}\)