Chapter 5.4, Problem 2E

To determine

(a)

To find:

The specified probability for the given scenario.

Answer to Problem 2E

Solution:

Value of the required probability is 0.2742.

Explanation of Solution

Formula used:

For a hyper geometric random variable $X$, the probability of obtaining $r$ successes in $n$ dependent trials is given by,

$P\left(X=r\right)=\frac{\left(C{}_r\right)\left(C{}_{n-r}\right)}{\left(C{}_n\right)}$

Here, $N$ is the number of items in the entire population,

$n$, is the number of trials,

$k$, is the number of successes in the entire population, and

$r$, is the number of success obtained in $n$ trials.

The formula to calculate $C{}_r$ is,

$C{}_r=\frac{k!}{r!\left(k-r\right)!}$

Calculation:

Total number of gift is 10.

$N=10$

Consider the event “gift is for Abby” as a success.

There are 5 gifts for Abby.

$k=5$

Number of trial is 4.

$n=4$.

Compute the probability for exactly 1 success.

Substitute 1 for $r$, 5 for $k$, 4 for $n$, and 10 for $N$ in the formula for exactly $r$ successes.

$\begin{array}{rll}P\left(X=1\right)& =& \frac{\left(C{}_1\right)\left(C{}_{4-1}\right)}{\left(C{}_4\right)}\\ & =& \frac{\left(C{}_1\right)\left(C{}_3\right)}{\left(C{}_4\right)}\\ & =& \frac{\frac{5!}{1!4!}\times \frac{5!}{3!2!}}{\frac{10!}{4!6!}}\\ & =& \frac{5\times 10}{210}\end{array}$

Further simplify the value of $P\left(X=1\right)$.

$\begin{array}{rll}P\left(X=1\right)& =& \frac{5\times 10}{210}\\ & =& 0.2380952381\end{array}$

$P\left(X=1\right)=0.2380$

Conclusion:

Thus, the probability of getting one gift for Abby is 0.2380.

To determine

(b)

To find:

The specified probability for the given scenario.

Answer to Problem 2E

Solution:

Value of the required probability is 0.8968.

Explanation of Solution

Formula used:

For a hyper geometric random variable $X$, the probability of obtaining at most $r$ successes in $n$ dependent trials is given by,

$P\left(X\le r\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+\mathrm{...}+P\left(X=r\right)$

$P\left(X=r\right)=\frac{\left(C{}_r\right)\left(C{}_{n-r}\right)}{\left(C{}_n\right)}$

Here, $N$ is the number of items in the entire population,

$n$, is the number of trials,

$k$, is the number of successes in the entire population, and

$r$, is the maximum number of successes obtained in $n$ trials.

The formula to calculate $C{}_r$ is,

$C{}_r=\frac{k!}{r!\left(k-r\right)!}$

Calculation:

Total number of gift is 10.

$N=10$

Consider the event “gift is for Andrew” as a success.

There are 5 gifts for Andrew.

$k=5$

Number of trial is 5.

$n=5$.

Compute the probability for at most 3 successes.

Substitute 3 for $r$ in the formula for at most $r$ successes.

$P\left(X\le 3\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)$

Substitute 0 for $r$, 5 for $k$, 5 for $n$, and 10 for $N$ in the formula for $P\left(X=r\right)$ to calculate $P\left(X=0\right)$.

$\begin{array}{rll}P\left(X=0\right)& =& \frac{\left(C{}_0\right)\left(C{}_{5-0}\right)}{\left(C{}_5\right)}\\ & =& \frac{\left(C{}_0\right)\left(C{}_5\right)}{\left(C{}_5\right)}\\ & =& \frac{\frac{5!}{0!5!}\times \frac{5!}{5!5!}}{\frac{10!}{5!5!}}\\ & =& \frac{1\times 1}{252}\end{array}$

Further simplify the value of $P\left(X=0\right)$.

$\begin{array}{rll}P\left(X=0\right)& =& \frac{1\times 1}{252}\\ & \approx & 0.003968\end{array}$

$P\left(X=0\right)=0.003968$

Substitute 1 for $r$, 5 for $k$, 5 for $n$, and 10 for $N$ in the formula for $P\left(X=r\right)$ to calculate $P\left(X=1\right)$.

$\begin{array}{rll}P\left(X=1\right)& =& \frac{\left(C{}_1\right)\left(C{}_{5-1}\right)}{\left(C{}_5\right)}\\ & =& \frac{\left(C{}_1\right)\left(C{}_4\right)}{\left(C{}_5\right)}\\ & =& \frac{\frac{5!}{1!4!}\times \frac{5!}{4!1!}}{\frac{10!}{5!5!}}\\ & =& \frac{5\times 5}{252}\end{array}$

Further simplify the value of $P\left(X=1\right)$.

$\begin{array}{rll}P\left(X=1\right)& =& \frac{25}{252}\\ & \approx & 0.099206\end{array}$

$P\left(X=1\right)=0.099206$

Substitute 2 for $r$, 5 for $k$, 5 for $n$, and 10 for $N$ in the formula for $P\left(X=r\right)$ to calculate $P\left(X=2\right)$.

$\begin{array}{rll}P\left(X=2\right)& =& \frac{\left(C{}_2\right)\left(C{}_{5-2}\right)}{\left(C{}_5\right)}\\ & =& \frac{\left(C{}_2\right)\left(C{}_3\right)}{\left(C{}_5\right)}\\ & =& \frac{\frac{5!}{2!3!}\times \frac{5!}{3!2!}}{\frac{10!}{5!5!}}\\ & =& \frac{10\times 10}{252}\end{array}$

Further simplify the value of $P\left(X=2\right)$.

$\begin{array}{rll}P\left(X=2\right)& =& \frac{100}{252}\\ & \approx & 0.3968254\end{array}$

$P\left(X=2\right)=0.3968254$

Substitute 3 for $r$, 5 for $k$, 5 for $n$, and 10 for $N$ in the formula for $P\left(X=r\right)$ to calculate $P\left(X=3\right)$.

$\begin{array}{rll}P\left(X=3\right)& =& \frac{\left(C{}_3\right)\left(C{}_{5-3}\right)}{\left(C{}_5\right)}\\ & =& \frac{\left(C{}_3\right)\left(C{}_2\right)}{\left(C{}_5\right)}\\ & =& \frac{\frac{5!}{3!2!}\times \frac{5!}{2!3!}}{\frac{10!}{5!5!}}\\ & =& \frac{10\times 10}{252}\end{array}$

Further simplify the value of $P\left(X=3\right)$.

$\begin{array}{rll}P\left(X=3\right)& =& \frac{100}{252}\\ & \approx & 0.3968254\end{array}$

$P\left(X=3\right)=0.3968254$

Add the values of $P\left(X=0\right),P\left(X=1\right),P\left(X=2\right),\mathrm{and}P\left(X=3\right)$ to get the value of $P\left(X\le 3\right)$.

$\begin{array}{rl}P\left(X\le 3\right)& =P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)\\ & =0.003968+0.099206+0.3968254+0.3968254\\ & \approx 0.8968\end{array}$

$P\left(X\le 3\right)=0.8968$

Conclusion:

Thus, the probability of getting at most 3 gifts for Andrew is 0.8968.