W04 Homework B

Due date: Tuesday 2/10, 11:59pm

Bernoulli process

01 $★$

08

Prize on the Mall

A booth on the Mall is running a secret prize game, in which the $5^{\text{th}}$ passerby wearing a hat wins $1,000.

Passersby wear hats independently of each other and with probability 20%.

Let $N$ be a random variable counting how many passersby pass by before a winner is found.

(a) What is the name of the distribution for $N$? What are the parameters?

(b) What is the probability that the ${10}^{\text{th}}$ passerby wins the prize?

(c) What is the probability that at least $7$ passersby are needed before a winner is found?

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Solution

Solutions - 5110-08

(a)

$L$ follows a Pascal distribution with parameters $p=0.2$ and $\ell =5$.

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(b)

Compute $P_L(10)$:

$$\begin{array}{c}{P}_L(10)=\left(\genfrac{}{}{0px}{}{k-1}{\ell -1}\right){(1-p)}^{k-\ell }{p}^{\ell }\\ \\ \gg \gg \left(\genfrac{}{}{0px}{}{9}{4}\right){(0.8)}^5{(0.2)}^5\gg \gg \approx 0.0132\end{array}$$

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(c)

Compute $P[L\ge 7]$:

The minimum number of passersby before a winner is declared is $5$, so $P[L\ge 7]=1-P_L(5)-P_L(6)$.

$$\begin{array}{c}P[L\ge 7]=1-P_L(5)-P_L(6)\\ \\ \gg \gg 1-\left(\genfrac{}{}{0px}{}{4}{4}\right){(0.8)}^0{(0.2)}^5-\left(\genfrac{}{}{0px}{}{5}{4}\right){(0.8)}^1{(0.2)}^5\gg \gg 0.9984\end{array}$$Link to original

02

06

Geometric distribution is memoryless

Suppose that $X\sim \text{Geom}(p)$. Derive this equation:

$$P[X=n+k|X>n]=P[X=k]$$

Interpret the equation. (Inspired by the title.)

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Solution

Solutions - 5110-06

(1) Set up conditional probability formula:

$$P[X=n+k|X>n]=\frac{P[X=n+k\cap X>n]}{P[X>n]}=\frac{P[X=n+k]}{P[X>n]}$$

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(2) Find formulas for numerator and denominator:

$$P[X=n+k]=p{(1-p)}^{n+k-1}$$ $$P[X>n]=\sum _{x=n+1}^{\infty }{(1-p)}^{x-1}p=\frac{p{(1-p)}^n}{1-(1-p)}={(1-p)}^n$$

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(3) Plug in and simplify:

$$P[X=n+k|X>n]=\frac{p{(1-p)}^{n+k-1}}{{(1-p)}^n}=p{(1-p)}^{k-1}=P[X=k]$$Link to original

03 $★$

07

Half the babies are female?

At Grace Community Hospital, 4 babies are delivered in one day. At Hope Valley Hospital, 6 babies are delivered in one day.

Consider these two events:

• (i) exactly half the babies born at Grace Community are female
• (ii) exactly half the babies born at Hope Valley are female

Perform a calculation to determine whether Event (i) is more probable, Event (ii) is more probable, or they are equally probable. (Assume the probability of each baby being born male or female is $0.5$.)

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Solution

Solutions - 5100-07

(i):

$$P[X=2]=\left(\genfrac{}{}{0px}{}{4}{2}\right){(0.5)}^2{(0.5)}^2=0.375$$

(ii):

$$P[X=3]=\left(\genfrac{}{}{0px}{}{6}{3}\right){(0.5)}^3{(0.5)}^3=0.3125$$

So event (i) is more probable.

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04

05

A very strange car

A very strange car with $n$ components will drive if at least half of its components work. Each component will work with the same probability $p$, independently of the others.

For what values of $p$ is a car with $n=3$ more likely to drive than a car with $n=5$?

(Start by defining a random variable that counts the number of working components.)

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Solution

Solutions - 5110-05

(1) Define random variables:

Let $X_3\sim \text{Bin}(3,p)$ represent the car with three components.

Let $X_5\sim \text{Bin}(5,p)$ represent the car with five components.

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(2) Find probabilities that both cars work:

For the three-component car, we want $P[X_3\ge 2]$, so

$$P[X_3\ge 2]=\sum _{k=2}^3\left(\genfrac{}{}{0px}{}{3}{k}\right)p^k{(1-p)}^{3-k}\gg \gg 3p^2-2p^3$$

For the five-component car, we want $P[X_5\ge 3]$, so

$$P[X_5\ge 3]=\sum _{k=3}^5\left(\genfrac{}{}{0px}{}{5}{k}\right)p^k{(1-p)}^{5-k}\gg \gg 10p^3-15p^4+6p^5$$

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(3) Find when $P[X_3\ge 2]>P[X_5\ge 3]$:

Solve the inequality for $p$:

$$\begin{array}{cc}3p^2-2p^3& >10p^3-15p^4+6p^5\\ 6p^5-15p^4+12p^3-3p^2& <0\\ 3p^2(2p^3-5p^2+4p-1)& <0\\ 3p^2{(p-1)}^2(2p-1)& <0\gg \gg 0<p<\frac{1}{2}\end{array}$$Link to original

Expectation and variance

05

04

Tutoring needs

A course with 6 students offers free one-on-one tutoring to each student for 1 hour the week before the final exam. One tutor, Jim, has been hired to provide this tutoring, but he is available for only 4 hours that week. The instructor of the course will tutor any students that Jim is not able to help. Jim will be paid $20 per hour by the department. The instructor will provide tutoring for free. Let $X$ be the number of students that will need tutoring. The PMF of $X$ is given below.

$$P_X(x)=\{\begin{array}{cc}0.10& x=0\\ 0.20& x=\mathrm{1,2}\\ 0.15& x=\mathrm{3,4,5}\\ 0.05& x=6\\ 0& \text{ otherwise }\end{array}$$

(a) Find the probability the instructor will need to provide tutoring.

(b) Find the expected value of the number of students that will need tutoring.

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Solution

Solutions - 5120-04

(a)

$$\begin{array}{c}P[X=5]+P[X=6]\gg \gg 0.15+0.05\gg \gg 0.20\end{array}$$

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(b)

$$\begin{array}{c}E[X]\gg \gg 0(0.10)+1(0.20)+2(0.20)+3(0.15)+4(0.15)+5(0.15)+6(0.05)\\ \\ \gg \gg 2.7\end{array}$$Link to original

06 $★$

01

Students and buses expect different crowding

Bus One has 10 students, Bus Two has 20, Bus Three has 30, and Bus Four has 40.

• Let $X$ measure the number of students on a given random student’s bus.
• Let $Y$ measure the number of students on a given random driver’s bus.

Compute $E[X]$ and $E[Y]$. Are they different? Why or why not?

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Solution

Solutions - 5120-01

(1) Compute $E[X]$:

There are $100$ total students.

Let $P[B_k]$ be the probability that Bus $k$ is selected, where $k\in \{1,2,3,4\}$.

Note that $P[B_k]=\frac{k}{10}$.

$$\begin{array}{c}E[X]=\sum _{k=1}^{4}P[B_k](10k)\\ \\ \gg \gg \sum _{k=1}^4{k}^2\gg \gg 30\end{array}$$

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(2) Compute $E[Y]$:

Let $P[B_k]$ be the probability that Bus $k$ is selected.

Since it’s based off the drivers, $P[B_k]=\frac{1}{4}$ for all $k$.

$$\begin{array}{c}E[Y]=\sum _{k=1}^{4}P[B_k]10k\\ \\ \gg \gg \frac{5}{2}\sum _{k=1}^{4}k\gg \gg 25\end{array}$$

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(3) Interpret solution:

For a driver, each bus is equally likely to be experienced.

For a student, you are more likely to find yourself on the bus with more students, so your experience is weighted towards the more crowded bus.

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Review problems

07

04

Rolling two dice

Two dice are rolled. Find the probabilities of the following events:

• $A$, the event that the sum is 10
• $B$, the event that the sum is 12
• $C$, the event that the two numbers are equal

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Solution

Solutions - 5070-04

There are $36$ total outcomes when rolling two dice.

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(1) Compute $P[A]$:

Desired outcomes for sum $=10$: $(4,6)$, $(5,5)$, $(6,4)$ — 3 outcomes.

$$P[A]=\frac{3}{36}=\frac{1}{12}$$

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(2) Compute $P[B]$:

Desired outcomes for sum $=12$: $(6,6)$ — 1 outcome.

$$P[B]=\frac{1}{36}$$

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(3) Compute $P[C]$:

Desired outcomes for equal numbers: $(\mathrm{1,1})$, $(\mathrm{2,2})$, $(\mathrm{3,3})$, $(\mathrm{4,4})$, $(\mathrm{5,5})$, $(\mathrm{6,6})$ — 6 outcomes.

$$P[C]=\frac{6}{36}=\frac{1}{6}$$Link to original

08 $★$

03

Bayes’ Theorem - DNA evidence

A crime is committed in a town of 100,000 citizens. After all 100,000 citizens’ DNA is analyzed, your friend Jim is found to have a DNA match to evidence at the scene. A forensics expert says that the probability of an innocent person matching this evidence is 0.01% (i.e. the false positive rate). How likely is it that Jim is guilty?

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Solution

Solutions - 5040-03

(1) Define events:

Let $G$ be the event that Jim is guilty.

Let $M$ be the event that the DNA matches.

We are given that $P[M|G^c]=0.0001$.

We assume that $P[M|G]=1$.

Since there are 100,000 citizens, $P[G]=0.00001$, $P[G^c]=0.99999$.

We are asked to compute $P[G|M]$.

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(2) Set up Bayes’ Theorem:

$$\begin{array}{c}P[G|M]=P[M|G]\frac{P[G]}{P[M]}\\ \\ \gg \gg \frac{P[M|G]P[G]}{P[M|G]P[G]+P[M|G^c]P[G^c]}\end{array}$$

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(3) Plug in values:

$$\begin{array}{c}P[G|M]=\frac{P[M|G]P[G]}{P[M|G]P[G]+P[M|G^c]P[G^c]}\\ \\ \gg \gg \frac{1\cdot 0.00001}{1\cdot 0.00001+0.0001\cdot 0.99999}\gg \gg \approx 0.0909\end{array}$$Link to original

Optional challenge problems

• Complete any one of these problems instead of two of the above problems of your choice.
• You must indicate on your paper which of the above two should be replaced by this one. Indicate at both the above two problems (can otherwise leave blank) to direct the grader to the last page where you work your choice of challenge problem.
• Limit of one substitution.

09

07

Binomial ratios

Suppose $X\sim \text{Bin}(n,p)$.

• Find the value of $k$ that maximizes $P_X(k)$. Do this by studying the successive ratios $P_X(k)/P_X(k-1)$.
• Use these ratios to compute $P[X\le 4]$ as a sum of 5 terms without using factorials. Do this by computing $P_X(0)$ directly, and then writing a recursive algorithm that determines $P_X(k)$ in terms of $P_X(k-1)$.

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Solution

Solutions - 5110-07

(1) Find formula for ratio $P_X(k)/P_X(k-1)$:

$$\begin{array}{cc}\frac{P_X(k)}{P_X(k-1)}& =\frac{\left(\genfrac{}{}{0px}{}{n}{k}\right)p^k{(1-p)}^{n-k}}{\left(\genfrac{}{}{0px}{}{n}{k-1}\right)p^{k-1}{(1-p)}^{n-(k-1)}}\\ & =\frac{\frac{n!}{k!(n-k)!}p}{\frac{n!}{(k-1)!(n-k+1)!}(1-p)}\\ & =\frac{(n-k+1)p}{k(1-p)}\end{array}$$

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(2) Interpret ratio:

We want $P_X(k)\ge P_X(k-1)$, so $(n-k+1)p\ge k(1-p)$.

Solving for $k$, we get $k\le (n+1)p$.

Since $k$ is an integer, $P_X(k)$ is maximized when $k=\lfloor (n+1)p\rfloor$.

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(3) Compute $P_X(0)$ directly:

Based on the figure, $n=10$ and $p=\frac{1}{2}$.

$$P_X(0)=\left(\genfrac{}{}{0px}{}{n}{0}\right)p^0{(1-p)}^n={(1-p)}^n={\left(\frac{1}{2}\right)}^{10}$$

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(4) Use ratio to solve for successive terms:

$$\begin{array}{cc}\frac{P_X(1)}{P_X(0)}& =\frac{np}{(1-p)}\\ P_X(1)& =\frac{5}{\frac{1}{2}}{\left(\frac{1}{2}\right)}^{10}& =5{\left(\frac{1}{2}\right)}^9\\ & \\ P_X(2)& =\frac{5}{\frac{1}{2}}\cdot 5{\left(\frac{1}{2}\right)}^9& =25{\left(\frac{1}{2}\right)}^8\\ & \\ P_X(3)& =\frac{5}{\frac{1}{2}}\cdot 25{\left(\frac{1}{2}\right)}^8& =125{\left(\frac{1}{2}\right)}^7\\ & \\ P_X(4)& =\frac{5}{\frac{1}{2}}\cdot 125{\left(\frac{1}{2}\right)}^7& =625{\left(\frac{1}{2}\right)}^6\end{array}$$

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(5) Add up probabilities:

$$P[X\le 4]=\sum _{k=0}^4{P}_X(k)\approx 0.3770$$Link to original

10

03

Expectation, variance of geometric variable

Derive formulas for $E[X]$ and $\mathrm{Var}[X]$ given $X\sim \mathrm{Geom}(p)$.

Hint: For $E[X]$ you will get a sum that has terms like $n{x}^{(n-1)}$.
This series comes from the geometric series $1/(1-x)=1+x+x^2+x^3+\dots$
(Differentiate both sides.)

For $E[X^2]$ you will need to consider this general fact of algebra: $A^2=A\cdot (A-1)+A$ (And apply the same methods as above.)

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Solution

Solutions - 5120-03

(1) State the PMF of a geometric random variable:

$$P_X(k)={(1-p)}^{k-1}p$$

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(2) Use formula for expectation to find $E[X]$:

$$\begin{array}{cc}E[X]& =\sum _{k=1}^{\infty }k{(1-p)}^{k-1}p\\ & =p\sum _{k=1}^{\infty }k{(1-p)}^{k-1}\end{array}$$

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(3) Apply hint:

We have that $\frac{1}{1-x}={\sum }_{k=0}^{\infty }{x}^k$.

Differentiating both sides yields $\frac{1}{{(1-x)}^2}={\sum }_{k=1}^{\infty }k{x}^{k-1}$.

Note that here, $x=(1-p)$, so

$$E[X]=p\frac{1}{{(1-(1-p))}^2}=\frac{p}{p^2}=\frac{1}{p}$$

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(4) Find expression for $E[X^2]$:

Note that $\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2$.

Applying the hint, we have $E[X^2]=E[X(X-1)+X]$.

Using the linearity of expectation, we can write this as $E[X(X-1)]+E[X]$.

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(5) Find $E[X(X-1)]$ and $E[X^2]$:

First, note that the second derivative of $\frac{1}{1-x}$ is $\frac{2}{{(1-x)}^3}={\sum }_{k=2}^{\infty }k(k-1)x^{k-2}$.

$$E[X(X-1)]=\sum _{k=1}^{\infty }k(k-1){(1-p)}^{k-1}p=\frac{2-p}{p^2}-\frac{1}{p}$$

Thus, $E[X^2]=\frac{2-p}{p^2}-\frac{1}{p}+\frac{1}{p}=\frac{2-p}{p^2}$.

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(6) Find $\text{Var}[X]$:

$$\text{Var}[X]=\frac{2-p}{p^2}-\frac{1}{p^2}=\frac{1-p}{p^2}$$Link to original