Chapter 13.6, Problem 36E

(a)

To determine

To find: Thenumbers of trials and the probability on each trials with meaning of the success in this case.

Answer to Problem 36E

The number of trials is $6$ and the probability of success on each trial is $20\%$ which is $0.2$ .

Explanation of Solution

Given: In one incident $6$ missiles are fired at fuel storage installation.

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

A success means that a missile hit its target. there are $6$ trials and the probability of success on each trial is $20\%$ which is $0.2$ .

Hence, the number of trials is $6$ and the probability of success on each trial is $20\%$ which is $0.2$ .

(b)

To determine

To find:the probability of hitting the target between $2$ and $6$ missiles.

Answer to Problem 36E

The probability of hitting the target between $2$ and $6$ is $\frac{1077}{3125}$ .

Explanation of Solution

Given: In one incident $6$ missiles are fired at fuel storage installation.

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

According to the given:

In one incident $6$ missiles are fired at fuel storage installation.

  $\begin{array}{l}P\left(\text{one misile hit the target}\right)=0.2=\frac{1}{5}.\\ P\left(\text{one misile dont hit the target}\right)=1-\frac{1}{5}=\frac{4}{5}.\end{array}$

The probability of hitting the target between $2$ and $6$ is:

  $\begin{array}{l}P\left(2\text{ to }6\text{ missile hit the target}\right)=1-\left(P\left(0\text{ missile hit the target}\right)+P\left(1\text{ missile hit the target}\right)\right).\\ \Rightarrow 1-\left(C{}_0\times {\left(\frac{1}{5}\right)}^0\times {\left(\frac{4}{5}\right)}^6+C{}_1\times {\left(\frac{1}{5}\right)}^1\times {\left(\frac{4}{5}\right)}^5\right).\\ \Rightarrow 1-\left(1\times 1\times \frac{4096}{15625}+6\times \frac{1}{5}\times \frac{1024}{3125}\right).\\ \Rightarrow \frac{1077}{3125}.\end{array}$

Hence, the probability of hitting the target between $2$ and $6$ is $\frac{1077}{3125}$ .