W04 - Examples

Bernoulli process

Binomial variable counting ones in repeated die rolls

Binomial variable counting ones in repeated die rolls

A standard die is rolled 6 times. Use a binomial variable to find the probability of rolling at least 4 ones.

Solution

(1) Labels.

Let $S\sim \text{Bin}(6,\frac{1}{6})$.

Interpret: $S$ counts the ones appearing over 6 rolls.

We seek $P[4\le S]$.

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(2) Calculations.

Exclusive events:

$$\begin{array}{c}P[4\le S]\gg \gg P_S(4)+P_S(5)+P_S(6)\\ \\ \gg \gg \left(\genfrac{}{}{0ex}{}{6}{4}\right){\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^2+\left(\genfrac{}{}{0ex}{}{6}{5}\right){\left(\frac{1}{6}\right)}^5{\left(\frac{5}{6}\right)}^1+\left(\genfrac{}{}{0ex}{}{6}{6}\right){\left(\frac{1}{6}\right)}^6{\left(\frac{5}{6}\right)}^0\\ \\ \gg \gg \frac{203}{23328}\gg \gg \approx 0.00870\end{array}$$ Link to original

Roll die until

Roll die until

Roll a fair die repeatedly. Find the probabilities that:

(a) At most 2 threes occur in the first 5 rolls.

(b) There is no three in the first 4 rolls, using a geometric variable.

Solution

(a)

(1) Label variables and events:

Use a variable $S\sim \mathrm{Bin}(5,1/6)$ to count the number of threes among the first six rolls.

Seek $P[S\le 2]$ as the answer.

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(2) Calculations:

Divide into exclusive events:

$$\begin{array}{c}P[S\le 2]\gg \gg P_S(0)+P_S(1)+P_S(2)\\ \\ \gg \gg \left(\genfrac{}{}{0ex}{}{5}{0}\right){\left(\frac{1}{6}\right)}^0{\left(\frac{5}{6}\right)}^5+\left(\genfrac{}{}{0ex}{}{5}{1}\right){\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^4+\left(\genfrac{}{}{0ex}{}{5}{2}\right){\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^3\\ \\ \gg \gg \frac{625}{648}\gg \gg \approx 0.965\end{array}$$

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(b)

(1) Label variables and events:

Use a variable $N\sim \mathrm{Geom}(1/6)$ to give the roll number of the first time a three is rolled.

Seek $P[N>4]$ as the answer.

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(2) Compute:

Sum the PMF formula for $\mathrm{Geom}(1/6)$:

$$P[N>4]\gg \gg \sum _{k=5}^{\infty }{\left(\frac{5}{6}\right)}^{k-1}\left(\frac{1}{6}\right)$$

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(3) Recall geometric series formula:

For any geometric series:

$$a+ar+a{r}^2+a{r}^3+\cdots =\frac{a}{1-r}$$

Therefore:

$$P[N>4]=\sum _{k=5}^{\infty }{\left(\frac{5}{6}\right)}^{k-1}\left(\frac{1}{6}\right)\gg \gg {\left(\frac{5}{6}\right)}^4$$ Link to original

Cubs winning the World Series

Cubs winning the World Series

Suppose the Cubs are playing the Yankees for the World Series. The first team to 4 wins in 7 games wins the series. What is the probability that the Cubs win the series?

Assume that for any given game the probability of the Cubs winning is $p=45\%$ and losing is $q=55\%$.

Solution

Method (a): We solve the problem using a binomial distribution.

(1) Label variables and events:

Use a variable $X\sim \mathrm{Bin}(7,p)$. This $X$ counts the number of wins over 7 games. Thus, for example, $P_X(4)$ is the probability that the Cubs win exactly 4 games over 7 played.

Seek $P_X(4)+P_X(5)+P_X(6)+P_X(7)$ as the answer.

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(2) Calculate using binomial PMF:

$$P_X(k)=\left(\genfrac{}{}{0ex}{}{7}{k}\right)p^k{q}^{7-k}$$

Insert data:

$$\begin{array}{c}{P}_X(4)+\cdots +P_X(7)\\ \\ \gg \gg \left(\genfrac{}{}{0ex}{}{7}{4}\right)p^4{q}^3+\left(\genfrac{}{}{0ex}{}{7}{5}\right)p^5{q}^2+\left(\genfrac{}{}{0ex}{}{7}{6}\right)p^6{q}^1+\left(\genfrac{}{}{0ex}{}{7}{7}\right)p^7{q}^0\end{array}$$

Compute:

$$\begin{array}{c}\gg \gg \frac{7\cdot 6\cdot 5}{3\cdot 2}{p}^4{q}^3+\frac{7\cdot 6}{2}{p}^5{q}^2+\frac{7}{1}{p}^6{q}^1+1\cdot p^7{q}^0\\ \\ \gg \gg p^4(35q^3+21p^1{q}^2+7p^{2}q+p^3)\end{array}$$

Convert $q\gg (1-p)$:

$$\begin{array}{c}\gg \gg p^4(35(1-p{)}^3+21p(1-p{)}^2+7p^2(1-p)+p^3)\\ \\ \gg \gg 35p^4-84p^5+70p^6-20p^7\gg \gg \approx 0.39\end{array}$$

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Method (b): We solve the problem using a Pascal distribution instead.

(1) Label variables and events:

Use a variable $Y\sim \mathrm{Pasc}(4,p)$. This $Y$ measures the discrete wait time until the $4^{\text{th}}$ win. Thus, for example, $P_Y(k)$ is the probability that the Cubs win their $4^{\text{th}}$ game on game number $k$.

Seek $P_Y(4)+P_Y(5)+P_Y(6)+P_Y(7)$ as the answer.

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(2) Calculate using Pascal PMF:

$$P_Y(k)=\left(\genfrac{}{}{0ex}{}{k-1}{3}\right)p^4{q}^{k-4}$$

Insert data:

$$\begin{array}{c}{P}_Y(4)+\cdots +P_Y(7)\\ \\ \gg \gg \left(\genfrac{}{}{0ex}{}{3}{3}\right)p^4{q}^0+\left(\genfrac{}{}{0ex}{}{4}{3}\right)p^4{q}^1+\left(\genfrac{}{}{0ex}{}{5}{3}\right)p^4{q}^2+\left(\genfrac{}{}{0ex}{}{6}{3}\right)p^4{q}^3\end{array}$$

Compute:

$$\begin{array}{c}\gg \gg 1\cdot p^4+\frac{4}{1}\cdot p^4{q}^1+\frac{5\cdot 4}{2}{p}^4{q}^2+\frac{6\cdot 5\cdot 4}{3\cdot 2}{p}^4{q}^3\\ \\ \gg \gg p^4(1+4q+10q^2+20q^3)\end{array}$$

Convert $q\gg (1-p)$:

$$\begin{array}{c}\gg \gg p^4(1+4(1-p)+10(1-p{)}^2+20(1-p{)}^3)\\ \\ \gg \gg 35p^4-84p^5+70p^6-20p^7\gg \gg \approx 0.39\end{array}$$

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Notice: The calculation seems very different than method (a), right up to the end!

Link to original

Expectation and variance

Gambling game - tokens in bins

Gambling game - tokens in bins

Consider a game like this: a coin is flipped; if $H$ then draw a token from Bin 1, if $T$ then from Bin 2.

• Bin 1 contents: 1 token $1,000, and 9 tokens $1
• Bin 2 contents: 5 tokens $50, and 5 tokens $1

It costs $50 to enter the game. Should you play it? (A lot of times?) How much would you pay to play?

Solution

(1) Setup:

Let $X$ be a random variable measuring your winnings in the game.

The possible values of $X$ are 1, 50, and 1000.

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(2) Find the PDF of $X$:

For $k=1$ have $P_X(1)\gg \gg \frac{1}{2}\cdot \frac{9}{10}+\frac{1}{2}\cdot \frac{5}{10}\gg \gg \frac{7}{10}$

For $k=50$ have $P_X(50)\gg \gg \frac{1}{2}\cdot \frac{5}{10}\gg \gg \frac{1}{4}$

For $k=1000$ have $P_X(1000)\gg \gg \frac{1}{2}\cdot \frac{1}{10}\gg \gg \frac{1}{20}$

These add to 1, and $P_X(x)=0$ for all other $x$.

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(3) Find $E[X]$ using the discrete formula:

$$\begin{array}{c}E[X]=\sum _{k}k\cdot P_X(k)\gg \gg 1\cdot P_X(1)+50\cdot P_X(50)+1000\cdot P_X(1000)\\ \\ \gg \gg 1\cdot \frac{7}{10}+50\cdot \frac{1}{4}+1000\cdot \frac{1}{20}\gg \gg \approx 63.2\end{array}$$

Since $63.2>50$, if you play it a lot at $50 you will generally make money.

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Challenge Q:
If you start with $200 and keep playing to infinity, how likely is it that you go broke?

Link to original

Expected value - rolling dice

Expected value: rolling dice

Let $X$ be a random variable counting the number of dots given by rolling a single die.

Then:

$$E[X]\gg \gg 1\cdot \frac{1}{6}+2\cdot \frac{1}{6}+\cdots +6\cdot \frac{1}{6}\gg \gg \frac{7}{2}$$

Let $S$ be an RV that counts the dots on a roll of two dice.

The PMF of $S$:

Then:

$$E[S]\gg \gg 2\cdot \frac{1}{36}+3\cdot \frac{2}{36}+4\cdot \frac{3}{36}+\cdots +12\cdot \frac{1}{36}\gg \gg 7$$

Notice that $\frac{7}{2}+\frac{7}{2}=7$.

In general, $E[X+Y]=E[X]+E[Y]$.

Let $X$ be a green die and $Y$ a red die.

From the earlier calculation, $E[X]=\frac{7}{2}$ and $E[Y]=\frac{7}{2}$.

Since $S=X+Y$, we derive $E[S]=7$ by simple addition!

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Expectation from PMF of related variable

Expected value by finding new PMF

Let $X$ have distribution given by this PMF:

Find $E[|X-2|]$.

Solution

(1) Compute the PMF of $|X-2|$:

PMF arranged by possible value:

$$\begin{array}{rl}P[|X-2|=0]& \gg \gg P[X=2]=\frac{1}{14}\\ P[|X-2|=1]& \gg \gg P[X=1]+P[X=3]=\frac{1}{7}+\frac{3}{14}=\frac{5}{14}\\ P[|X-2|=2]& \gg \gg P[X=4]=\frac{2}{7}\\ P[|X-2|=3]& \gg \gg P[X=5]=\frac{2}{7}\\ P[|X-2|=k]& \gg \gg 0\text{for}k\ne 0,1,2,3.\end{array}$$

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(2) Calculate the expectation:

Using formula for discrete PMF:

$$E[|X-2|]\gg \gg 0\cdot \frac{1}{14}+1\cdot \frac{5}{14}+2\cdot \frac{2}{7}+3\cdot \frac{2}{7}\gg \gg \frac{25}{14}$$ Link to original

Variance for composite using PMF and simpler formula

Variance for composite using PMF and simpler formula

Suppose $X$ has this PMF:

$k:$	1	2	3
$P_X(k):$	$1/7$	$2/7$	$4/7$

Find $\mathrm{Var}\left[\frac{1}{1+X}\right]$ using the formula $\mathrm{Var}[Y]=E[Y^2]-E[Y{]}^2$ with $Y=\frac{1}{1+X}$.

(Hint: you should find $E[Y]=\frac{13}{42}$ and $E[Y^2]=\frac{13}{126}$ along the way.)

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Poisson process

Poisson calculation

Poisson calculation

Suppose $X\sim \mathrm{Pois}(10)$. Find $P[X\le 13\mid X\ge 7]$. (Leave the answer in exact form.)

Solution

(1) Conditioning definition:

$$P[X\le 13\mid X\ge 7]\gg \gg \frac{P[7\le X\le 13]}{1-P[X<7]}$$

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(2) Expand numerator:

$$P[7\le X\le 13]\gg \gg e^{-10}\frac{10^7}{7!}+e^{-10}\frac{10^8}{8!}+\cdots +e^{-10}\frac{10^{13}}{13!}$$

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(3) Simplify:

$$\gg \gg e^{-10}\frac{10^7}{7!}(1+\frac{10}{8}+\frac{10^2}{8\cdot 9}+\cdots +\frac{10^6}{8\cdot 9\cdots 13})$$

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(4) Compute for denominator:

$$P[X<7]\gg \gg e^{-10}\frac{10^0}{0!}+e^{-10}\frac{10^1}{1!}+\cdots +e^{-10}\frac{10^6}{6!}$$ Link to original

Arrivals at a post office

Arrivals at a post office

Client arrivals at a post office are modelled well using a Poisson variable.

Each potential client has a very low and independent chance of coming to the post office, but there are many thousands of potential clients, so the arrivals at the office actually come in moderate number.

Suppose the average rate is 5 clients per hour.

(a) Find the probability that nobody comes in the first 10 minutes of opening. (The cashier is considering being late by 10 minutes to run an errand on the way to work.)

(b) Find the probability that 5 clients come in the first hour. (I.e. the average is achieved.)

(c) Find the probability that 9 clients come in the first two hours.

Solution

(a)

Expect $5/6$ clients every 10 minutes. Therefore, let $X\sim \mathrm{Pois}(5/6)$. Seek $P_X(0)$ as the answer.

PMF:

$$P_X(k)=e^{-5/6}\frac{(5/6{)}^k}{k!}$$

Insert data and compute:

$$P_X(0)\gg \gg e^{-5/6}\gg \gg \approx 0.435$$

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(b)

Rate is already correct. Let $X\sim \mathrm{Pois}(5)$.

PMF:

$$P_X(5)=e^{-5}\frac{5^5}{5!}\gg \gg \approx 0.175$$

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(c)

Expect 10 clients every 2 hours. Therefore let $X\sim \mathrm{Pois}(10)$.

PMF:

$$P_X(9)\gg \gg e^{-10}\frac{10^9}{9!}\gg \gg \approx 0.125$$

Notice that 0.125 is smaller than 0.175.

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