Examples

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:

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

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 $S\sim \mathrm{Bin}(\mathrm{5,1}/6)$ 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:

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

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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:

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

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:

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

Insert data:

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

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:

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

Insert data:

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

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!