Chapter 4.3, Problem 52E

a.

To determine

Compute the value of $P\left(A\right)$.

Answer to Problem 52E

$P\left(A\right)=0.3$.

Explanation of Solution

Calculation:

There were a total of 1,000 components, 300 of them are defective. Two components are selected at random. Event A denotes that first component selected is defective and B denote that second component selected is defective.

The probability of the event A can be obtained by the formula:

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Number of outcomes in}\text{sample space}}$

Here, event A denotes that first component selected is defective. Among the 1,000 components, 300 of them are defective.

Substitute 300 for “number of outcomes in A” and 1,000 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(A\right)=\frac{300}{1,000}\\ =0.3\end{array}$

Thus, $P\left(A\right)=0.3$.

b.

To determine

Compute the value of $P\left(B|A\right)$.

Answer to Problem 52E

$P\left(B|A\right)=0.2993$.

Explanation of Solution

Calculation:

The probability of the event $B|A$  can be obtained by the formula:

$P\left(B|A\right)=\frac{\text{Number of outcomes in }\left(B\text{|}A\right)}{\text{Number of outcomes in}\text{sample space}}$

Event A denote that first component selected is defective and B denote that second component selected is defective.

Here, it is given that first component selected is defective, therefore the number of remaining defectives in the lot is 299 $\left(=300-1\right)$ and the total number of components in the lots becomes 999$\left(=1,000-1\right)$. Second defective component is selected from the remaining 999 components containing 299 defectives.

Substitute 299 for “number of outcomes in $B|A$” and 999 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(B|A\right)=\frac{299}{999}\\ =0.29929\\ =0.2993\end{array}$

Thus, $P\left(B|A\right)=0.2993$.

c.

To determine

Compute the value of $P\left(A\text{and }B\right)$.

Answer to Problem 52E

$P\left(A\text{and }B\right)=0.0898$.

Explanation of Solution

Calculation:

The formula for conditional probability is by:

$P\left(B|A\right)=\frac{P\left(A\text{and}B\right)}{P\left(A\right)}$

Therefore,

$P\left(A\text{and}B\right)=P\left(B|A\right)\times P\left(A\right)$

From part (a), $P\left(A\right)=\frac{300}{1,000}$.

From part (b), $P\left(B|A\right)=\frac{299}{999}$.

Therefore,

$\begin{array}{c}P\left(A\text{ and }B\right)=\frac{300}{1,000}\times \frac{299}{999}\\ =\frac{299}{3330}\\ =0.08978\\ \approx 0.0898\end{array}$

Thus, $P\left(A\text{and }B\right)=0.0898$.

d.

To determine

Check whether events A and B are independent or not. Is it reasonable to treat A and B as though they were independent. Explain the reason.

Answer to Problem 52E

Events A and B are not independent.

Explanation of Solution

Event A denote that first component selected is defective and B denote that second component selected is defective.

Independent events:

Two events are said to be independent if occurrence of one does not affect the occurrence of other. The thumb rule states that, if the sample size is less than 5% of the population size then sample items are considered to be independent.

Event that second component drawn is defective depend on whether or not the first components defective. Therefore, events A and B are depends on each other.

Therefore, events A and B are not independent.

Here, there were a total of 1,000 components; two components are selected at random. Therefore, the sample size is 2 and population size is 1,000.

$\begin{array}{c}\text{5\% of population size}=\left(5\%\right)\times \left(1,000\right)\\ =\frac{5}{100}\times 1,000\\ =50\end{array}$

That is, sample size $\left(=2\right)$ is less than 5% of the population size $\left(=50\right)$. Therefore, it is reasonable to treat A and B as though they were independent.