Chapter 5, Problem 10P

a.

To determine

Check whether $A_1\text{and }{A}_2$ are mutually exclusive or not.

 Explain the answer.

Answer to Problem 10P

Events $A_1\text{and }{A}_2$ are mutually exclusive.

Explanation of Solution

Calculation:

The prior probabilities for the events $A_1\text{and }{A}_2$ are$P\left(A_1\right)=0.40\text{and }P\left(A_2\right)=0.60$.

The values of $P\left(A_1\cap A_2\right)=0,P\left(B|A_1\right)=0.20,\text{and }P\left(B|A_2\right)=0.05$ .

Mutually exclusive events:

Two or more events are said to be mutually exclusive if they have no element in common. For mutually exclusive events, $P\left(A\cap B\right)=0$.

Here, $P\left(A_1\cap A_2\right)=0$. Therefore events $A_1\text{and }{A}_2$ are mutually exclusive.

b.

To determine

Compute the probability values$P\left(A_1\cap B\right)\text{and}P\left(A_2\cap B\right)$.

Answer to Problem 10P

The probabilities are:

$\underset{¯}{P\left(A_1\cap B\right)=0.08},\underset{¯}{P\left(A_2\cap B\right)=0.03}.$

Explanation of Solution

Calculation:

The Multiplication Rule:

For any two events A and B, the multiplication rule states that $P\left(A\cap B\right)=P\left(B|A\right)\times P\left(A\right)$.

Required probabilities are:

$\begin{array}{l}P\left(A_1\cap B\right)=P\left(B|A_1\right)\times P\left(A_1\right)\\ P\left(A_2\cap B\right)=P\left(B|A_2\right)\times P\left(A_2\right)\end{array}$

Here, $P\left(A_1\right)=0.40,P\left(A_2\right)=0.60,P\left(B|A_1\right)=0.20$$\text{and }P\left(B|A_2\right)=0.05$.

Substitute these values in the above formula.

Therefore,

$\begin{array}{c}P\left(A_1\cap B\right)=\left(0.20\right)\times \left(0.40\right)\\ =0.08\end{array}$

Thus, $P\left(A_1\cap B\right)=0.08$.

Therefore,

$\begin{array}{c}P\left(A_2\cap B\right)=\left(0.05\right)\times \left(0.60\right)\\ =0.03\end{array}$

Thus, $P\left(A_2\cap B\right)=0.03$.

c.

To determine

Compute the probability value$P\left(B\right)$.

Answer to Problem 10P

The probability value is, $\underset{¯}{P\left(B\right)=0.11}$.

Explanation of Solution

Calculation:

The probability of the event can be obtained as follows:

$P\left(B\right)=P\left(A_1\cap B\right)+P\left(A_2\cap B\right)$.

From part (b), $P\left(A_1\cap B\right)=0.08\text{and }P\left(A_2\cap B\right)=0.03$.

Substitute these values in the above formula.

Therefore,

$\begin{array}{c}P\left(B\right)=0.08+0.03\\ =0.11\end{array}$

Thus, $\underset{¯}{P\left(B\right)=0.11}$.

d.

To determine

Compute $P\left(A_1|B\right)\text{and }P\left(A_2|B\right)$ using Bayes’ theorem.

Answer to Problem 10P

The probability values are:

$\underset{¯}{P\left(A_1|B\right)=0.7273},\underset{¯}{P\left(A_2|B\right)=0.2727}.$

Explanation of Solution

Calculation:

Bayes Theorem (Two-Event Case):

$\begin{array}{c}P\left(A_1|B\right)=\frac{P\left(A_1\right)\times P\left(B|A_1\right)}{P\left(A_1\right)\times P\left(B|A_1\right)+P\left(A_2\right)\times P\left(B|A_2\right)}\\ =\frac{P\left(A_1\right)\times P\left(B|A_1\right)}{P\left(B\right)}\\ P\left(A_2|B\right)=\frac{P\left(A_2\right)\times P\left(B|A_2\right)}{P\left(A_1\right)\times P\left(B|A_1\right)+P\left(A_2\right)\times P\left(B|A_2\right)}\\ =\frac{P\left(A_2\right)\times P\left(B|A_2\right)}{P\left(B\right)}\end{array}$

From part (b), $P\left(A_1\cap B\right)=0.08\text{and }P\left(A_2\cap B\right)=0.03$.

From part (c), $P\left(B\right)=0.11$.

Substitute these values in the Bayes’ theorem.

Therefore,

$\begin{array}{c}P\left(A_1|B\right)=\frac{0.08}{0.11}\\ =0.72727\\ =0.7373\end{array}$

Thus, $\underset{¯}{P\left(A_1|B\right)=0.7273}$.

$\begin{array}{c}P\left(A_2|B\right)=\frac{0.03}{0.11}\\ =0.2727\end{array}$

Thus, $\underset{¯}{P\left(A_2|B\right)=0.2727}$.