Chapter 5.2, Problem 21E

To determine

(a)

To find:

The specified probability for the given scenario.

Answer to Problem 21E

Solution:

Value of the required probability is 1.

Explanation of Solution

Formula used:

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

$\begin{array}{rllllll}P\left(X\ge r\right)& =& P\left(X=r\right)+P(X& =& r+1)+\mathrm{...}+P(X& =& n)\\ & =& 1-P\left(X\le r-1\right)\end{array}$

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

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, $r$ 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 “surviving of plant” as a success.

The probability of any plant surviving in Kerry’s garden is 0.8.

Probability of success is $0.8$.

$p=0.8$

Number of trial is 19.

$n=19$.

Compute the probability for at least 5 successes.

Substitute 5 for $r$ in the formula for at least $r$ successes.

$P\left(X\ge 5\right)=1-P\left(X\le 4\right)$

$P\left(X\le 4\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)$

Substitute 19 for $n$, 0 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=0\right)$.

$\begin{array}{rl}P\left(X=0\right)& =\frac{19!}{0!19!}\times {\left(0.8\right)}^0\times {\left(1-0.8\right)}^{19}\\ & =1\times 1\times {\left(0.2\right)}^{19}\\ & \approx 0\end{array}$

Substitute 19 for $n$, 1 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=1\right)$.

$\begin{array}{rll}P\left(X=1\right)& =& \frac{19!}{1!18!}\times {\left(0.8\right)}^1\times {\left(1-0.8\right)}^{18}\\ & =& 19\times \left(0.8\right)\times {\left(0.2\right)}^{18}\\ & \approx & 0\end{array}$

Substitute 19 for $n$, 2 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=2\right)$.

$\begin{array}{rll}P\left(X=2\right)& =& \frac{19!}{2!17!}\times {\left(0.8\right)}^2\times {\left(1-0.8\right)}^{17}\\ & =& 19\times 9\times {\left(0.8\right)}^2\times {\left(0.2\right)}^{17}\\ & \approx & 0\end{array}$

Substitute 19 for $n$, 3 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=3\right)$.

$\begin{array}{rll}P\left(X=3\right)& =& \frac{19!}{3!16!}\times {\left(0.8\right)}^3\times {\left(1-0.8\right)}^{16}\\ & =& 19\times 3\times 17\times {\left(0.8\right)}^3\times {\left(0.2\right)}^{16}\\ & \approx & 0\end{array}$

Substitute 19 for $n$, 4 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=4\right)$.

$\begin{array}{rll}P\left(X=4\right)& =& \frac{19!}{4!15!}\times {\left(0.8\right)}^4\times {\left(1-0.8\right)}^{15}\\ & =& 19\times 17\times 4\times {\left(0.8\right)}^4\times {\left(0.2\right)}^{15}\\ & \approx & 0\end{array}$

Add all the values of

$P\left(X=0\right),P\left(X=1\right),P\left(X=2\right),P\left(X=3\right),\text{and}P\left(X=4\right)$

$\begin{array}{l}P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)\\ =0+0+0+0+0\\ =0\end{array}$

$P\left(X\le 4\right)=0$

Substitute 0 for $P\left(X\le 4\right)$ in expansion for $P\left(X\ge 5\right)$.

$\begin{array}{rll}P\left(X\ge 5\right)& =& 1-0\\ & =& 0\end{array}$

$P\left(X\ge 5\right)=1$

Conclusion:

Thus, the probability of surviving at least 5 of them is 1.

To determine

(b)

To find:

The specified probability for the given scenario.

Answer to Problem 21E

Solution:

Value of the required probability is 0.6733.

Explanation of Solution

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 “surviving of plant” as a success.

The probability of any plant surviving in Kerry’s garden is 0.8.

Probability of success is $0.8$.

$p=0.8$

Number of trial is 19.

$n=19$.

Required number of success is $\frac{3}{4}\text{of}19$.

$\begin{array}{rll}r& =& \frac{3}{4}\times 19\\ & =& 14.25\end{array}$

$r$ is a positive integer.

Take $r=14$

Compute the probability for more than 14 successes.

Substitute 14 for $r$, 19 for $n$ in the formula for more than $r$ successes.

$P\left(X>14\right)=P\left(X=15\right)+P\left(X=16\right)+P\left(X=17\right)+P\left(X=18\right)+P\left(X=19\right)$

Substitute 19 for $n$, 15 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=15\right)$

$\begin{array}{rll}P\left(X=15\right)& =& \frac{19!}{15!4!}\times {\left(0.8\right)}^{15}\times {\left(1-0.8\right)}^4\\ & =& 3876\times {\left(0.8\right)}^{15}\times {\left(0.2\right)}^4\\ & =& 0.2181994019\end{array}$

Substitute 19 for $n$, 16 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=16\right)$

$\begin{array}{rll}P\left(X=16\right)& =& \frac{19!}{16!3!}\times {\left(0.8\right)}^{16}\times {\left(1-0.8\right)}^3\\ & =& 969\times {\left(0.8\right)}^{16}\times {\left(0.2\right)}^3\\ & =& 0.2181994019\end{array}$

Substitute 19 for $n$, 17 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=17\right)$

$\begin{array}{rll}P\left(X=17\right)& =& \frac{19!}{17!2!}\times {\left(0.8\right)}^{17}\times {\left(1-0.8\right)}^2\\ & =& 171\times {\left(0.8\right)}^{17}\times {\left(0.2\right)}^2\\ & =& 0.1540231073\end{array}$

Substitute 19 for $n$, 18 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=18\right)$

$\begin{array}{rll}P\left(X=18\right)& =& \frac{19!}{18!1!}\times {\left(0.8\right)}^{18}\times {\left(1-0.8\right)}^1\\ & =& 19\times {\left(0.8\right)}^{18}\times {\left(0.2\right)}^1\\ & =& 0.06845471434\end{array}$

Substitute 19 for $n$, 19 for $r$, and 0.8 for $p$ in the formula of $P\left(X=r\right)$ to calculate $P\left(X=19\right)$

$\begin{array}{rll}P\left(X=19\right)& =& \frac{19!}{19!0!}\times {\left(0.8\right)}^{19}\times {\left(1-0.8\right)}^0\\ & =& 1\times {\left(0.8\right)}^{19}\times 1\\ & =& 0.01441151881\end{array}$

Add all the values of

$P\left(X=15\right),P\left(X=16\right),P\left(X=17\right),P\left(X=18\right),\text{and}P\left(X=19\right)$

$\begin{array}{l}P\left(X=15\right)+P\left(X=16\right)+P\left(X=17\right)+P\left(X=18\right)+P\left(X=19\right)\\ =0.2181994019+0.2181994019+0.1540231073+0.06845471434+0.01441151881\\ \approx 0.6733\end{array}$

$P\left(X>14\right)=0.6733$

Conclusion:

Thus, the probability of surviving more than $\frac{3}{4}$ of them is 0.6733.

To determine

(c)

To find:

The specified probability for the given scenario.

Answer to Problem 21E

Solution:

Value of the required probability is 0

Explanation of Solution

Formula used:

For a binomial random variable $X$, the probability of obtaining no successes 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 “surviving of plant” as a success.

The probability of any plant surviving in Kerry’s garden is 0.8.

Probability of success is $0.8$.

$p=0.8$

Number of trial is 19.

$n=19$.

Compute the probability for no successes.

Substitute 19 for $n$, 0 for $r$, and 0.8 for $p$ in the formula of $P\left(X=0\right)$

$\begin{array}{rll}P\left(X=0\right)& =& \frac{19!}{0!19!}\times {\left(0.8\right)}^0\times {\left(1-0.8\right)}^{19}\\ & =& 1\times 1\times {\left(0.2\right)}^{19}\\ & \approx & 0\end{array}$

Conclusion:

Thus, the probability that no plant survive is 0.