W03 Homework B

Due date: Tuesday 2/3, 11:59pm

Repeated trials

01 $★$

04

Guessing on a test

Your odds of getting any given exam question right are $80\%$. The exam has 4 questions, and you need to answer 3 correctly to pass.

(a) What is the probability that you pass?

(b) After finishing the exam, you are 100% sure that you got the second question right. Now what are the odds that you pass?

Link to original

Solution

Solutions - 5080-04

(a)

$$\begin{array}{c}P[\text{pass}]=\sum _{k=3}^4\left(\genfrac{}{}{0px}{}{4}{k}\right){(0.8)}^k{(0.2)}^{4-k}\\ \\ \gg \gg \approx 0.8192\end{array}$$

---

(b)

Out of the remaining three questions, you need to answer at least two correctly.

$$\begin{array}{c}P[\text{pass}]=\sum _{k=2}^3\left(\genfrac{}{}{0px}{}{3}{k}\right){(0.8)}^k{(0.2)}^{3-k}\\ \\ \gg \gg \approx 0.896\end{array}$$Link to original

02

01

Independent trials - At least 45 good paper clips

For a paper clip production line, 90% of the paper clips come off good, and 10% come off broken.

You buy a box of 50 paper clips from this line. What is the probability that at least 45 of them are good?

Link to original

Solution

Solutions - 5080-01

(1) Identify the binomial distribution:

$X\sim \text{Bin}(50,0.9)$ where $X$ = number of good paper clips.

---

(2) Express and evaluate $P[X\ge 45]$:

$$\begin{array}{c}P[X\ge 45]=\sum _{k=45}^{50}\left(\genfrac{}{}{0px}{}{50}{k}\right){(0.9)}^k{(0.1)}^{50-k}\\ \\ \gg \gg \approx 0.6161\end{array}$$Link to original

Reliability

03

01

Reliability for complex process

Consider a process with the following diagram of components in series and parallel:

Use $W_i$ to denote the event that component $i$ succeeds.

Suppose the success probabilities per component are given by this chart:

1	2	3	4	5	6	7	8
80%	60%	40%	90%	80%	50%	70%	90%

What are the odds of success for the whole process?

Link to original

Solution

Solutions - 5090-01

(1) Find the probability components 4 and 5 both succeed (denote $S_1$):

$$P[S_1]=P[W_4]\cdot P[W_5]=0.9\cdot 0.8=0.72$$

---

(2) Find the probability the parallel block of 3 and $S_1$ succeeds (denote $S_2$):

$$P[S_2]=1-P[S_1^c]P[W_3^c]=1-0.28\cdot 0.6=0.832$$

---

(3) Find the probability the series 2, $S_2$, 6 succeeds (denote $S_3$):

$$P[S_3]=P[W_2]P[S_2]P[W_6]=0.6\cdot 0.832\cdot 0.5=0.2496$$

---

(4) Find the probability the parallel block of $S_3$ and 7 succeeds (denote $S_4$):

$$P[S_4]=1-P[S_3^c]P[W_7^c]=1-0.7504\cdot 0.3=0.77488$$

---

(5) Find the probability the full system 1, $S_4$, 8 succeeds (denote $S$):

$$\begin{array}{c}P[S]=P[W_1]P[S_4]P[W_8]\\ \\ \gg \gg 0.8\cdot 0.77488\cdot 0.9\gg \gg \approx 0.5579\end{array}$$Link to original

Discrete random variables

04 $★$

03

PMF from CDF for tracking gasoline

I am trying to keep track of how much gasoline (measured in gallons, rounded to the nearest integer, and denoted by $X$) my car uses every week. I have managed to find the CDF of $X$:

$$F(x)=\{\begin{array}{cc}0& \phantom{0\le }x<0\\ 0.1& 0\le x<1\\ 0.2& 1\le x<2\\ 0.5& 2\le x<3\\ 1& x\ge 3\end{array}$$

Assume gasoline costs $3 a gallon. Let $Y$ denote the amount of money I spend on gasoline every week. What is the PMF of $Y$?

Link to original

Solution

Solutions - 5100-03

(1) Find the PMF of $X$ from the CDF:

$k$	0	1	2	3
$P_X(k)$	0.1	0.1	0.3	0.5

---

(2) Find the PMF of $Y=3X$:

Since $Y=3X$, the possible values of $Y$ are $3\cdot 0=0$, $3\cdot 1=3$, $3\cdot 2=6$, $3\cdot 3=9$, each with the same probability as the corresponding $X$ value.

$k$	0	3	6	9
$P_Y(k)$	0.1	0.1	0.3	0.5

Link to original

05

02

Gambling with a coin

Two players, A and B, are flipping a fair coin together. If it comes up heads, A pays $1 to B, and if it comes up tails, B pays $1 to A.

They play five rounds. Let $X$ be a random variable recording A’s final winnings.

(a) Find the set of possible values of $X$. (I.e., the set of outcomes with nonzero probability.)

(b) Find the PMF and CDF of $X$.

Link to original

Solution

Solutions - 5100-02

(a)

Let $X_i$ be A’s winnings after the $i^{\text{th}}$ round. Since A either loses or gains one dollar each round, $X_i\in \{\pm 1\}$. We have:

$$X=\sum _{i=1}^5{X}_i$$

Therefore $X\in \{-5,-3,-1,1,3,5\}$.

---

(b)

(1) Describe the PMF of $X$:

$X=5$ if and only if all 5 rounds are tails, so $P[X=5]=\frac{1}{2^5}=\frac{1}{32}$.

$X=-5$ if and only if all 5 rounds are heads, so $P[X=-5]=\frac{1}{32}$.

$X=3$ if 4 rounds are tails, and $X=-3$ if 4 rounds are heads. Since the coin is fair, $P[X=3]=P[X=-3]=\left(\genfrac{}{}{0px}{}{5}{4}\right){(0.5)}^5=\frac{5}{32}$.

$X=1$ if 3 rounds are tails, and $X=-1$ if 3 rounds are heads. Since the coin is fair, $P[X=1]=P[X=-1]=\left(\genfrac{}{}{0px}{}{5}{3}\right){(0.5)}^5=\frac{10}{32}$.

---

(2) Define the PMF of $X$:

$$P_X(x)=\{\begin{array}{cc}\frac{1}{32}& x=-5\\ \frac{5}{32}& x=-3\\ \frac{10}{32}& x=-1\\ \frac{10}{32}& x=1\\ \frac{5}{32}& x=3\\ \frac{1}{32}& x=5\\ 0& \text{otherwise}\end{array}$$

---

(3) Define the CDF of $X$:

Note that the jumps in the CDF occur at the possible discrete values of $X$.

$$F_X(x)=\{\begin{array}{cc}0& x<-5\\ \frac{1}{32}& -5\le x<-3\\ \frac{6}{32}& -3\le x<-1\\ \frac{16}{32}& -1\le x<1\\ \frac{26}{32}& 1\le x<3\\ \frac{31}{32}& 3\le x<5\\ 1& 5\le x\end{array}$$Link to original

06 $★$

01

Digit of a real number

Suppose a real number is chosen randomly in the unit interval $[\mathrm{0,1}]$. Consider the decimal expansion of this number. Let $Y$ be a random variable giving the first digit after the decimal point. Find the possible values, the PMF, and the CDF of $Y$.

Link to original

Solution

Solutions - 5100-01

(1) Find the possible values of $Y$:

After the decimal point, any of the 10 digits can appear, so $Y\in \{0,1,\dots ,9\}$.

---

(2) Find the PMF of $Y$:

Each digit has a $\frac{1}{10}$ chance of appearing.

$$P_Y(y)=\{\begin{array}{cc}\frac{1}{10}& y\in \{0,1,\dots ,9\}\\ 0& \text{otherwise}\end{array}$$

---

(3) Find the CDF of $Y$:

$$F_Y(y)=\{\begin{array}{cc}0& y<0\\ \frac{1}{10}(y+1)& 0\le y\le 9\\ 1& 9<y\end{array}$$Link to original

Review problems

07

06

Multinomial - Colored marbles in a line

How many ways are there to line up 10 colored marbles (2 red, 3 white, 5 blue), assuming you cannot distinguish marbles of the same color?

Link to original

Solution

Solutions - 5070-06

This is a standard multinomial coefficient application.

Number the marbles 1, 2, … 10 and then count the ways of putting them into three bins: red bin size 2, white bin size 3, blue bin size 5.

In other words, for the three bins: $r_1=2$, $r_2=3$, and $r_3=5$.

$$\begin{array}{c}\left(\genfrac{}{}{0px}{}{n}{r_1,r_2,r_3}\right)\gg \gg \left(\genfrac{}{}{0px}{}{10}{\mathrm{2,3,5}}\right)\\ \\ \gg \gg \frac{10!}{2!3!5!}\gg \gg 2520\end{array}$$Link to original

08

07

Multiplication Rule - Fund performance

The odds of the Winning Fund outperforming the market in a random year are 15%. The odds that it outperforms the market in a 1-year period assuming it has done so in the prior year are 30%.

What is the probability of the Winning Fund outperforming the market in 2 consecutive years?

Link to original

Solution

Solutions - 5030-07

(1) Define events:

Let $W_1$ be the event where the Winning Fund outperforms the first year.

Let $W_2$ be the event where the Winning Fund outperforms the second year.

We are asked to compute $P[W_1\cap W_2]$.

---

(2) Identify given probabilities:

$P[W_1]=0.15$

$P[W_2|W_1]=0.3$

---

(3) Apply the multiplication rule:

$$\begin{array}{c}P[W_1\cap W_2]=P[W_1]\cdot P[W_2|W_1]\\ \\ \gg \gg 0.15\cdot 0.3\gg \gg 0.045\end{array}$$Link to original

09

08

Applicant qualifications A

A hiring manager will randomly select two people from a group of 5 applicants. Of the 5 applicants, 2 are more qualified and 3 are less qualified (but the manager does not know this).

If at least one of the less qualified applicants is selected, what is the probability that both applicants selected will be less qualified?

Link to original

Solution

Solutions - 5030-08

(1) Define events and count outcomes:

Let $A$ = both selected are less qualified, and $B$ = at least one less qualified is selected.

Total ways to choose 2 from 5: $\left(\genfrac{}{}{0px}{}{5}{2}\right)=10$

---

(2) Compute $P[A]$ and $P[B]$:

Ways both are less qualified: $\left(\genfrac{}{}{0px}{}{3}{2}\right)=3$, so $P[A]=\frac{3}{10}$

Ways neither is less qualified (both more qualified): $\left(\genfrac{}{}{0px}{}{2}{2}\right)=1$, so $P[B^c]=\frac{1}{10}$ and $P[B]=\frac{9}{10}$

Since $A\subset B$, we have $P[A\cap B]=P[A]=\frac{3}{10}$.

---

(3) Apply conditional probability:

$$\begin{array}{c}P[A|B]=\frac{P[A\cap B]}{P[B]}\\ \\ \gg \gg \frac{3/10}{9/10}\gg \gg \frac{1}{3}\end{array}$$Link to original

10

03

Applicant qualifications B

A hiring manager will randomly select two people from a group of 5 applicants. Of the 5 applicants, 2 are more qualified and 3 are less qualified (but the manager does not know this).

Let Event A be selecting 2 more qualified applicants and Event B be selecting 2 less qualified applicants. Determine whether A and B are independent events and justify your answer.

Link to original

Solution

Solutions - 5050-03

(1) Compute $P[A]$ and $P[B]$:

Total ways to choose 2 from 5: $\left(\genfrac{}{}{0px}{}{5}{2}\right)=10$

$$P[A]=\frac{\left(\genfrac{}{}{0px}{}{2}{2}\right)}{\left(\genfrac{}{}{0px}{}{5}{2}\right)}=\frac{1}{10},P[B]=\frac{\left(\genfrac{}{}{0px}{}{3}{2}\right)}{\left(\genfrac{}{}{0px}{}{5}{2}\right)}=\frac{3}{10}$$

---

(2) Check independence:

$P[A\cap B]=0$ since selecting 2 people cannot yield both “2 more qualified” and “2 less qualified” simultaneously.

$$P[A\cap B]=0\ne \frac{1}{10}\cdot \frac{3}{10}=P[A]\cdot P[B]$$

So $A$ and $B$ are not independent.

Link to original