Chapter 2.3, Problem 24E

A lot of 1000 components contains 300 that are defective. Two components are drawn at random and tested. Let A be the event that the first component drawn is defective, and let B be the event that the second component drawn is defective.

1. a. Find P(A).
2. b. Find P(B|A).
3. c. Find P(A Ç B).
4. d. Find P(A^c Ç B).
5. e. Find P(B).
6. f. Find P(A|B).
7. g. Are A and B independent? Is it reasonable to treat A and B as though they were independent? Explain.

To determine

Find $P\left(A\right)$.

Answer to Problem 24E

The $P\left(A\right)$ is $\frac{3}{10}$.

Explanation of Solution

Given info:

A lot of 1,000 components contain 300 that are defective. Two components are drawn at random and tested. Let A be the event that the first component drawn is defective and let B be the event that the second component drawn is defective.

Calculation:

From given information, it can be observed that A be the event that the first component drawn is defective.

The number of defective components is 300 and total number of components is 1,000.

The formula for probability of the event that the first component drawn is defective is,

$P\left(A\right)=\frac{\text{Number of defective components}}{\text{Total number of components}}$

That is,

$\begin{array}{c}P\left(A\right)=\frac{300}{1,000}\\ =\frac{3}{10}\end{array}$

Thus, $P\left(A\right)$ is $\frac{3}{10}$.

To determine

Find $P\left(B|A\right)$.

Answer to Problem 24E

The $P\left(B|A\right)$ is $\frac{299}{999}$.

Explanation of Solution

Calculation:

From given information, it can be observed that B be the event that the second component drawn is defective.

The number possible ways of the first component drawn is defective is 300 out of 1,000 (n₁). Therefore, the number of ways of getting the second one is defective is 299 $\left(=300-1\right)$ out of 999 $\left(=1,000-1\right)$(n₂).

The probability that second one selected is defective given that the first one was defective $P\left(B|A\right)$ is obtained as,

$P\left(B|A\right)=\frac{\left(\begin{array}{l}\text{Number of ways of getting }\\ \text{the second one is defective}\end{array}\right)}{n_2}$

Substitute 299 for “Number of ways of getting second one is defective” and 999 for n₂

$P\left(B|A\right)=\frac{299}{999}$

Thus, the $P\left(B|A\right)$ is $\frac{299}{999}$.

To determine

Find $P\left(A\cap B\right)$.

Answer to Problem 24E

The $P\left(A\cap B\right)$ is $\frac{299}{3330}$.

Explanation of Solution

Calculation:

The probability that both are defective $P\left(A\cap B\right)$ is obtained by multiplying probability that first one selected is defective and probability that there is second one selected is defective given that first one was defective.

That is,

$P\left(A\cap B\right)=P\left(A\right)P\left(B|A\right)$

Substitute $P\left(A\right)=\frac{3}{10}$ and $P\left(B|A\right)=\frac{299}{999}$

$\begin{array}{c}P\left(A\cap B\right)=\frac{3}{10}\times \frac{299}{999}\\ =\frac{299}{3330}\end{array}$

Thus, the $P\left(A\cap B\right)$ is $\frac{299}{3330}$.

To determine

Find $P\left(A^c\cap B\right)$.

Answer to Problem 24E

The $P\left(A^c\cap B\right)$ is $\frac{70}{333}$.

Explanation of Solution

Calculation:

The probability $P\left(A^c\cap B\right)$ is obtained by multiplying probability that first one selected is not defective and probability that there is second one selected is defective given that first one was not defective.

That is,

$P\left(A^c\cap B\right)=P\left(A^c\right)P\left(B|A^c\right)$

A^c be the event that the first component drawn is not defective.

The formula for finding $P\left(A^c\right)$ is,

$P\left(A^c\right)=1-P\left(A\right)$

Substitute $P\left(A\right)=\frac{3}{10}$

$\begin{array}{c}P\left(A^c\right)=1-\frac{3}{10}\\ =\frac{7}{10}\end{array}$

The number possible ways of the first component drawn is not defective is 1,000 out of 1,000 (n₁). Therefore, the number of ways of getting the second one is defective is 300 out of 999 $\left(=1,000-1\right)$(n₂).

The probability that second one selected is defective given that the first one was not defective $P\left(B{|A}^c\right)$ is obtained as,

$P\left(B{|A}^c\right)=\frac{\left(\begin{array}{l}\text{Number of ways of getting }\\ \text{the second one is defective}\end{array}\right)}{n_2}$

Substitute 300 for “Number of ways of getting second one is defective” and 999 for n₂

$P\left(B{|A}^c\right)=\frac{300}{999}$

Thus, the $P\left(B{|A}^c\right)$ is $\frac{3}{9}$.

Substitute $P\left(A^c\right)=\frac{7}{10}$ and $P\left(B{|A}^c\right)=\frac{300}{999}$

$\begin{array}{c}P\left(A^c\cap B\right)=\frac{7}{10}\times \frac{300}{999}\\ =\frac{70}{333}\end{array}$

Thus, the $P\left(A^c\cap B\right)$ is $\frac{70}{333}$.

To determine

Find $P\left(B\right)$.

Answer to Problem 24E

The $P\left(B\right)$ is $\frac{3}{10}$.

Explanation of Solution

Calculation:

From part (c), $P\left(A\cap B\right)=\frac{299}{3330}$ and from part (d), $P\left(A^c\cap B\right)=\frac{70}{333}$.

The formula for finding $P\left(B\right)$ is,

$P\left(B\right)=P\left(A\cap B\right)+P\left(A^c\cap B\right)$

Substitute $P\left(A\cap B\right)=\frac{299}{3330}$ and $P\left(A^c\cap B\right)=\frac{70}{333}$

$\begin{array}{c}P\left(B\right)=\frac{299}{3330}+\frac{70}{333}\\ =\frac{999}{3330}\\ =\frac{3}{10}\end{array}$

Thus, the $P\left(B\right)$ is $\frac{3}{10}$.

To determine

Find $P\left(A|B\right)$.

Answer to Problem 24E

The $P\left(A|B\right)$ is $\frac{299}{999}$.

Explanation of Solution

Calculation:

Conditional probability:

If A and B are two events with $P\left(B\right)\ne 0$ then the conditional probability of A given B is,

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

Substitute $P\left(A\cap B\right)=\frac{299}{3330}$ and $P\left(B\right)=\frac{3}{10}$

$\begin{array}{c}P\left(A|B\right)=\frac{\frac{299}{3330}}{\frac{3}{10}}\\ =\frac{299}{999}\end{array}$

Thus, the $P\left(A|B\right)$ is $\frac{299}{999}$.

To determine

Explain whether A and B are independent or not also explain whether it is reasonable to treat A and B independent.

Explanation of Solution

Calculation:

From part (a), $P\left(A\right)=\frac{3}{10}$, part (c), $P\left(A\cap B\right)=\frac{299}{3330}$ and from part (e), $P\left(B\right)=\frac{3}{10}$.

Independent of events:

It the probability of occurrence of one event does not influence on the other event then the two events A and B are said to independent.

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

Substitute $P\left(A\cap B\right)=\frac{299}{3330}$, $P\left(A\right)=\frac{3}{10}$ and $P\left(B\right)=\frac{3}{10}$

$\begin{array}{c}P\left(A\cap B\right)=P\left(A\right)P\left(B\right)\\ \frac{299}{3330}=\frac{3}{10}\times \frac{3}{10}\\ \frac{299}{3330}=\frac{9}{100}\\ 0.0898\ne 0.09\end{array}$

Thus, A and B are not independent because $P\left(A\cap B\right)\ne P\left(A\right)P\left(B\right)$ but it is observed they are approximately equal. Hence it can be concluded that it is reasonable to treat A and B independent.