Chapter 8.6, Problem 59E

(a).

To determine

To find:

The probability that all four units are gold.

Answer to Problem 59E

The required probability is $0.255$ .

Explanation of Solution

Given:

A shipment of $12$ microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number of microwave ovens are $12$ .

Total number of good microwave ovens are $9$ .

Total number of defective microwave ovens are $3$ .

Total number of selection of $4$ from $12$ is $C{}_4=495$ .

Selection of $4$ good from $9$ good is $C{}_4=126$ .

Thus, probability of all $4$ are good is as follows:

  $\begin{array}{c}P\left(Good\right)=\frac{126}{495}\\ =0.255\end{array}$

Conclusion:

Hence, the required probability is $0.255$ .

(b).

To determine

To find:

The probability that exactly two units are good.

Answer to Problem 59E

The required probability is $0.2181$ .

Explanation of Solution

Given:

A shipment of $12$ microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number of microwave ovens are $12$ .

Total number of good microwave ovens are $9$ .

Total number of defective microwave ovens are $3$ .

Total number of selection of $4$ from $12$ is $C{}_4=495$ .

Selection of $2$ good from $9$ good is $C{}_2=36$ .

Selection of $2$ defective from $3$ defective is $C{}_2=3$ .

Thus, probability of exactly $2$ are good is as follows:

  $\begin{array}{c}P\left(Exactly2Good\right)=\frac{C{}_2\cdot C{}_2}{C{}_4}\\ =\frac{36\cdot 3}{495}\\ =0.2181\end{array}$

Conclusion:

Hence, the required probability is $0.2181$ .

(c).

To determine

To find:

The probability that at least two units are good.

Answer to Problem 59E

The required probability is $0.982$ .

Explanation of Solution

Given:

A shipment of $12$ microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number of microwave ovens are $12$ .

Total number of good microwave ovens are $9$ .

Total number of defective microwave ovens are $3$ .

Total number of selection of $4$ from $12$ is $C{}_4=495$ .

Selection of $2$ good from $9$ good is $C{}_2=36$ .

Selection of $2$ defective from $3$ defective is $C{}_2=3$ .

Selection of $3$ good from $9$ good is $C{}_3=84$ .

Selection of $1$ defective from $3$ defective is $C{}_1=3$ .

Thus, probability of at least $2$ are good is $P\left(2\text{ good}\right)+P\left(\text{3 good}\right)+P\left(\text{4 good}\right)$ .

  $\begin{array}{c}P\left(Exactly2Good\right)=\frac{C{}_2\cdot C{}_2}{C{}_4}+\frac{C{}_3\cdot C{}_1}{C{}_4}+\frac{C{}_4\cdot C{}_4}{C{}_4}\\ =\frac{108}{495}+\frac{252}{495}+\frac{126}{495}\\ =\frac{486}{495}\\ =0.982\end{array}$

Conclusion:

Hence, the required probability is $0.982$ .