Chapter 2.3, Problem 23E

A lot of 10 components contains 3 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. Are A and B independent? Explain.

To determine

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

Answer to Problem 23E

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

Explanation of Solution

Given info:

A lot of 10 components contain 3 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 3 and total number of components is 10.

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,

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

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

To determine

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

Answer to Problem 23E

The $P\left(B|A\right)$ is $\frac{2}{9}$.

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 3 out of 10 (n₁). Therefore, the number of ways of getting the second one is defective is 2 $\left(=3-1\right)$ out of 9 $\left(=10-1\right)$(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 2 for “Number of ways of getting second one is defective” and 9 for n₂

$P\left(B|A\right)=\frac{2}{9}$

Thus, the $P\left(B|A\right)$ is $\frac{2}{9}$.

To determine

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

Answer to Problem 23E

The $P\left(A\cap B\right)$ is $\frac{1}{15}$.

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 $P\left(B|A\right)=\frac{2}{9}$

$\begin{array}{c}P\left(A\cap B\right)=\frac{3}{10}\times \frac{2}{9}\\ =\frac{1}{15}\end{array}$

Thus, the $P\left(A\cap B\right)$ is $\frac{1}{15}$.

To determine

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

Answer to Problem 23E

The $P\left(A^c\cap B\right)$ is $\frac{7}{30}$.

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 10 out of 10 (n₁). Therefore, the number of ways of getting the second one is defective is 3 out of 9 $\left(=10-1\right)$(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 3 for “Number of ways of getting second one is defective” and 9 for n₂

$P\left(B{|A}^c\right)=\frac{3}{9}$

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

Substitute $P\left(A^c\right)=\frac{7}{10}$ and $P\left(B|A^c\right)=\frac{3}{9}$

$\begin{array}{c}P\left(A^c\cap B\right)=\frac{7}{10}\times \frac{3}{9}\\ =\frac{7}{30}\end{array}$

Thus, the $P\left(A^c\cap B\right)$ is $\frac{7}{30}$.

To determine

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

Answer to Problem 23E

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

Explanation of Solution

Calculation:

From part (c), $P\left(A\cap B\right)=\frac{1}{15}$ and from part (d), $P\left(A^c\cap B\right)=\frac{7}{30}$.

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 $P\left(A\cap B\right)=\frac{1}{15}$ and $P\left(A^c\cap B\right)=\frac{7}{30}$

$\begin{array}{c}P\left(B\right)=\frac{1}{15}+\frac{7}{30}\\ =\frac{3}{10}\end{array}$

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

To determine

Explain whether A and B are independent or not.

Answer to Problem 23E

No, A and B are not independent because $P\left(A\cap B\right)\ne P\left(A\right)P\left(B\right)$.

Explanation of Solution

Calculation:

From part (a), $P\left(A\right)=\frac{3}{10}$, part (c), $P\left(A\cap B\right)=\frac{1}{15}$ 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 $P\left(A\cap B\right)=\frac{1}{15}$, $P\left(A\right)=\frac{3}{10}$ and $P\left(B\right)=\frac{3}{10}$

$\begin{array}{c}P\left(A\cap B\right)=P\left(A\right)P\left(B\right)\\ \frac{1}{15}=\frac{3}{10}\times \frac{3}{10}\\ \frac{1}{15}\ne \frac{9}{100}\end{array}$

Thus, the A and B are not independent.