Quiz 5 Solution

The following formulas fold for all $\alpha$ with $|\alpha |<1$:

1. $1+\alpha +{\alpha }^2+{\alpha }^3+\cdots =\sum _{k=0}^{\mathrm{\infty }}{\alpha }^k=\frac{1}{1-\alpha }$
2. $1+2\alpha +3{\alpha }^2+4{\alpha }^3+\cdots =\sum _{k=1}^{\mathrm{\infty }}k{\alpha }^{k-1}=\frac{1}{(1-\alpha {)}^2}$

Question 1

A fair coin is tossed repeatedly until it comes up heads, and the number $N$ of tosses needed is recorded.  Each toss is independent of all other tosses.  Let $A$ be the event that $N$ is an even number. What is the conditional probability mass function (pmf) of $N$ given that $A$ occurred?  Hint: $Pr(A)=\frac{1}{3}$.

Solution:  Note that ${\mathcal{S}}_N=\{1,2,3,\dots \}$, and $A$ is the event that $N\in \{2,4,6,8,\dots \}$.   We have

$p_N(n\mid A)=Pr(N=n\mid A)=\frac{Pr(N\in (A\cap \{n\}))}{Pr(A)}$

$=\{\begin{array}{ll}\frac{p_N(n)}{1/3},& n\in \{2,4,6,\dots \}\\ 0,& \text{otherwise}\end{array}$

$=\{\begin{array}{ll}\frac{3}{2^n},& n\in \{2,4,6,\dots \}\\ 0,& \text{otherwise}\end{array}$

Question 2

An urn contains two green balls and one yellow ball.  In each trial of a random experiment, a ball is sampled from the urn, its colour is noted, and the ball is returned to the urn.  Different trials are independent.   Let $X$ denote the number of trials needed until the yellow ball is drawn from the urn.  The range of $X$ is then ${\mathcal{S}}_X=\{1,2,3,\dots \}$.    What is the expected value, $E(X)$,  of $X$?

Solution:  $X$ is a positive geometric random variable with parameter $p=\frac{1}{3}$, and so has expected value $E(X)=\frac{1}{p}=3$.

We can also compute the expected value explicitly (repeating an example from the lectures) as follows.  Note that $p_X(k)=p(1-p{)}^{k-1}$ with $p=\frac{1}{3}$, where $k\in \{1,2,3,\dots \}$.  Then

$E(X)=\sum _{k\in {\mathcal{S}}_X}k\cdot p_X(k)=\sum _{k=1}^{\mathrm{\infty }}kp(1-p{)}^{k-1}$

$=p\sum _{k=1}^{\mathrm{\infty }}k(1-p{)}^{k-1}$

$=p\cdot \frac{1}{(1-(1-p){)}^2}$ (this is one of the sums noted at the top)

$=\frac{p}{p^2}=\frac{1}{p}=3$.

Question 3

For the experiment of the previous question, what is the expected value of $X$ given that $X\ne 1$?  (In other words, given that the yellow ball isn't drawn from the urn on the very first trial, what is the expected number of trials needed until it actually is drawn?)

Solution:  Intuitively, we expect 3 additional trials following the first (unsuccessful) trial for a total of 4.

We can verify this explicitly as follows. We have, for $p=\frac{1}{3}$,

$p_X(k\mid X\ne 1)=Pr(X=k\mid X\ne 1)$

$=\frac{Pr(X=k\text{ and }X\ne 1)}{Pr(X\ne 1)}$

$=\{\begin{array}{ll}0,& k=1,\\ \frac{Pr(X=k)}{1-Pr(X=1)},& k>1\end{array}$

$=\{\begin{array}{ll}0,& k=1,\\ \frac{p(1-p{)}^{k-1}}{1-p},& k>1\end{array}$

$=\{\begin{array}{ll}0,& k=1,\\ p(1-p{)}^{k-2},& k>1.\end{array}$

Thus $E(X\mid X\ne 1)=\sum _{k\in {\mathcal{S}}_X}k\cdot p_X(k\mid X\ne 1)$

$=\sum _{k=2}^{\mathrm{\infty }}kp(1-p{)}^{k-2}$  (omitting $k=1$ term with value zero)

$=\frac{p}{1-p}(2(1-p)+3(1-p{)}^2+4(1-p{)}^3+\cdots )$ (this is the second sum at the top with the first term missing)

$=\frac{p}{1-p}(-1+1+2(1-p)+3(1-p{)}^2+4(1-p{)}^3+\cdots )$

$=\frac{p}{1-p}(-1+\frac{1}{(1-(1-p){)}^2})$

$=\frac{p}{1-p}(-1+\frac{1}{p^2})$

$=\frac{p}{1-p}\cdot \frac{1-p^2}{p^2}=\frac{1+p}{p}=1+\frac{1}{p}=4$.

Question 4

The number of patrons entering the library over the noon hour is a Poisson random variable with an expected value of 6.   What is the probability that fewer than 3 patrons enter the library over the noon hour?

Solution:  Let $X$ denote the number of patrons entering the library over the noon hour.  Since $X$ is a Poisson random variable with an expected value of 6, we have $p_X(k)=\exp (-6)\frac{6^k}{k!}$, for $k\in \{0,1,2,\dots \}$.  It follows that

$Pr(X<3)=\sum _{k=0}^2{p}_X(k)=e^{-6}(1+6+\frac{6^2}{2})=25\exp (-6)\approx 0.0619688$.

Question 5

For the same library as in the previous question, what is the probability that more than 9 patrons enter the library over the noon hour?

Solution:  With $X$ defined as in Question 4, $Pr(X>9)=1-Pr(X\le 9)=1-\exp (-6)(1+6+\frac{6^2}{2!}+\cdots +\frac{6^9}{9!})\approx 0.083924$.

Question 6

A data center has ${10}^5$ disk drives.  Suppose that disks fail independently on any given day with probability $2\cdot {10}^{-5}$.  Find the probability that there are no disk failures on a given day.

Solution:  The number of disk failures is has a binomial distribution with parameters $n={10}^5$ and $p=2\cdot {10}^{-5}$.   We use the Poisson approximation with parameter $\alpha =np=2$, and probability mass function $p_X(k)=\exp (-2)\frac{2^k}{k!}$, $k=0,1,2,\dots$.  Then $Pr(X=0)=p_X(0)=\exp (-2)\approx 0.135335$.

Question 7

For the same data center as in the previous question, find the probability that there are more than 4 disk failures on a given day.

Solution:  Using the same Poisson approximation, we have $Pr(X>4)=1-Pr(X\le 4)=1-\exp (-2)(1+2+\frac{2^2}{2}+\frac{2^3}{3!}+\frac{2^4}{4!})$

$=1-7\exp (-2)\approx 0.052653$