Chapter 5.2, Problem 25E

To determine

(a)

To find:

The specified probability for the given scenario.

Answer to Problem 25E

Solution:

Value of the required probability is 0.9952.

Explanation of Solution

Given information:

Ronnie owns a fireworks stand and knows that in the fireworks business, 1 out of every 13 fireworks is a dud. Suppose that Juanita buys 10 firecrackers at Ronnie’s stand.

Formula used:

For a binomial random variable $X$, the probability of obtaining no more than $r$ successes in $n$ independent trials is given by,

$1-P\left(X>r\right)=P\left(X\le r\right)$

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

The formula to calculate $P\left(X=r\right)$ is,

$P\left(X=r\right)=C{}_r\cdot p^r{\left(1-p\right)}^{\left(n-r\right)}$

Here, $r+1$ is the minimum number of successes,

$n$, is the number of trials, and

$p$, is the probability of getting a success on any trial.

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

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

Calculation:

Consider the event “duds” as a success.

Probability of winning is 1 out of 13.

Probability of success is $\frac{1}{13}$.

$p=\frac{1}{13}\approx 0.076$

Number of trial is 10.

$n=10$.

Compute the probability for no more than 3 successes.

Substitute 3 for $r$, in the formula for no more than $r$ successes.

$1-P\left(X>3\right)=P\left(X\le 3\right)$

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

Substitute 10 for $n$, 0 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=0\right)$

$\begin{array}{rll}P\left(X=0\right)& =& \frac{10!}{0!10!}\times {\left(0.076\right)}^0\times {\left(1-0.076\right)}^{10}\\ & =& 1\times 1\times {\left(0.924\right)}^{10}\\ & =& 0.4536487434\end{array}$

Substitute 10 for $n$, 1 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=1\right)$

$\begin{array}{rll}P\left(X=1\right)& =& \frac{10!}{1!9!}\times {\left(0.076\right)}^1\times {\left(1-0.076\right)}^9\\ & =& 10\times \left(0.076\right)\times {\left(0.924\right)}^9\\ & =& 0.373131001\end{array}$

Substitute 10 for $n$, 2 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=2\right)$

$\begin{array}{rll}P\left(X=2\right)& =& \frac{10!}{2!8!}\times {\left(0.076\right)}^2\times {\left(1-0.076\right)}^8\\ & =& 45\times {\left(0.076\right)}^2\times {\left(0.924\right)}^8\\ & =& 0.138106929\end{array}$

Substitute 10 for $n$, 3 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=3\right)$

$\begin{array}{rll}P\left(X=3\right)& =& \frac{10!}{3!7!}\times {\left(0.076\right)}^3\times {\left(1-0.076\right)}^7\\ & =& 120\times {\left(0.076\right)}^3\times {\left(0.924\right)}^7\\ & =& 0.03029185166\end{array}$

Add the values of $P\left(X=0\right),P\left(X=1\right),P\left(X=2\right),\text{and}P\left(X=3\right)$.

$\begin{array}{l}P\left(X\le 3\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)\\ =0.4536487434+0.373131001+0.138106929+0.03029185166\\ \approx 0.9952\end{array}$

Conclusion:

Thus, the probability that no more than 3 are duds is 0.9952.

To determine

(b)

To find:

The specified probability for the given scenario.

Answer to Problem 25E

Solution:

Value of the required probability is 0.45.

Explanation of Solution

Given information:

Ronnie owns a fireworks stand and knows that in the fireworks business, 1 out of every 13 fireworks is a dud. Suppose that Juanita buys 10 firecrackers at Ronnie’s stand.

Formula used:

For a binomial random variable $X$, the probability of no success in $n$ independent trials is given by,

$P\left(X=0\right)=C{}_0\cdot p^0{\left(1-p\right)}^{\left(n-0\right)}$

$n$, is the number of trials, and

$p$, is the probability of getting a success on any trial

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

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

Calculation:

Consider the event “duds” as a success.

Probability of winning is 1 out of 13.

Probability of success is $\frac{1}{13}$.

$p=\frac{1}{13}\approx 0.076$

Number of trial is 10.

$n=10$.

Compute the probability for no success.

Substitute 10 for $n$, 0 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=0\right)$

$\begin{array}{rll}P\left(X=0\right)& =& \frac{10!}{0!10!}\times {\left(0.076\right)}^0\times {\left(1-0.076\right)}^{10}\\ & =& 1\times 1\times {\left(0.924\right)}^{10}\\ & \approx & 0.45\end{array}$

$P\left(X=0\right)=0.45$

Conclusion:

Thus, the probability of no duds is 0.45.

(c)

To determine

To find:

The specified probability for the given scenario.

Answer to Problem 25E

Solution:

Value of the required probability is 0.00003.

Explanation of Solution

Given information:

Ronnie owns a fireworks stand and knows that in the fireworks business, 1 out of every 13 fireworks is a dud. Suppose that Juanita buys 10 firecrackers at Ronnie’s stand.

Formula used:

For a binomial random variable $X$, the probability of obtaining more than $r$ successes in $n$ independent trials is given by,

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

The formula to calculate $P\left(X=r\right)$ is,

$P\left(X=r\right)=C{}_r\cdot p^r{\left(1-p\right)}^{\left(n-r\right)}$

Here, $r+1$ is the minimum number of successes,

$n$, is the number of trials, and

$p$, is the probability of getting a success on any trial

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

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

Calculation:

Consider the event “duds” as a success.

Probability of winning is 1 out of 13.

Probability of success is $\frac{1}{13}$.

$p=\frac{1}{13}\approx 0.076$

Number of trial is 10.

$n=10$.

Consider number of more than half of the crackers as a number of success.

$r>\frac{10}{2}=5$

Compute the probability for more than 5 successes.

Substitute 5 for $r$, 10 for $n$ in the formula for more than $r$ successes.

$P\left(X>5\right)=P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right)+P\left(X=9\right)+P\left(X=10\right)$

Substitute 10 for $n$, 6 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=6\right)$

$\begin{array}{rll}P\left(X=6\right)& =& \frac{10!}{6!4!}\times {\left(0.076\right)}^6\times {\left(1-0.076\right)}^4\\ & =& 210\times {\left(0.076\right)}^6\times {\left(0.924\right)}^4\\ & =& 0.0000294977393\end{array}$

Substitute 10 for $n$, 7 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=7\right)$

$\begin{array}{rll}P\left(X=7\right)& =& \frac{10!}{7!3!}\times {\left(0.076\right)}^7\times {\left(1-0.076\right)}^3\\ & =& 120\times {\left(0.076\right)}^7\times {\left(0.924\right)}^3\\ & =& 0.00000138641199\end{array}$

Substitute 10 for $n$, 8 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=8\right)$

$\begin{array}{rll}P\left(X=8\right)& =& \frac{10!}{8!2!}\times {\left(0.076\right)}^8\times {\left(1-0.076\right)}^2\\ & =& 45\times {\left(0.076\right)}^8\times {\left(0.924\right)}^2\\ & \approx & 0\end{array}$

Substitute 10 for $n$, 9 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=9\right)$

$\begin{array}{rll}P\left(X=9\right)& =& \frac{10!}{9!1!}\times {\left(0.076\right)}^9\times {\left(1-0.076\right)}^1\\ & =& 10\times {\left(0.076\right)}^9\times {\left(0.924\right)}^1\\ & \approx & 0\end{array}$

Substitute 10 for $n$, 10 for $r$, and 0.076 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=10\right)$

$\begin{array}{rll}P\left(X=10\right)& =& \frac{10!}{10!0!}\times {\left(0.076\right)}^{10}\times {\left(1-0.076\right)}^0\\ & =& 1\times {\left(0.076\right)}^{10}\times {\left(0.924\right)}^0\\ & \approx & 0\end{array}$

Add all the values of

$P\left(X=6\right),P\left(X=7\right),P\left(X=8\right),P\left(X=9\right),\text{and}P\left(X=10\right)$

$\begin{array}{l}P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right)+P\left(X=9\right)+P\left(X=10\right)\\ =0.0000294977393+0.00000138641199+0+0+0\\ \approx 0.00003\end{array}$

$P\left(X\ge 10\right)=0.000016$

Conclusion:

Thus, the probability of more than half duds is 0.00003.