Chapter 7.3, Problem 7.54E

(a)

To determine

To find: the $P\left(B|M\right)$ and means of this value

Answer to Problem 7.54E

0.5938

Explanation of Solution

Given:

	Male	Female	Total
Eat breakfast regularly	190	110	300
Don’t eat breakfast regularly	130	165	295
Total	320	275	595

Calculation:

  $\begin{array}{c}P\left(B|M\right)=\frac{190}{190+130}\\ =\frac{190}{320}\\ =0.5938\end{array}$

There $P\left(B|M\right)$ is indicating that the probability of a male eats breakfast regularly.

(b)

To determine

To find: $P\left(M|B\right)$ and means of this value.

Answer to Problem 7.54E

0.6333

Explanation of Solution

Given:

	Male	Female	Total
Eat breakfast regularly	190	110	300
Don’t eat breakfast regularly	130	165	295
Total	320	275	595

Calculation:

  $\begin{array}{c}P\left(M|B\right)=\frac{190}{300}\\ =0.6333\end{array}$

There $P\left(M|B\right)$ is indicating that the probability that eats breakfast regularly is a male.

(c)

To determine

To Explain: that events B and M mutually exclusive or not, justify answer.

Answer to Problem 7.54E

Not mutually exclusive

Explanation of Solution

Given:

	Male	Female	Total
Eat breakfast regularly	190	110	300
Don’t eat breakfast regularly	130	165	295
Total	320	275	595

Calculation:

  $\begin{array}{c}P\left(M\cap B\right)=\frac{190}{595}\\ =0.3193\end{array}$

There $P\left(M\cap B\right)=0.3193$ this is not equal to zero which is indicating that the events are not mutually exclusive.

(d)

To determine

To Explain: that the events B and M independent or not, justify answer.

Answer to Problem 7.54E

Not independent

Explanation of Solution

Given:

	Male	Female	Total
Eat breakfast regularly	190	110	300
Don’t eat breakfast regularly	130	165	295
Total	320	275	595

Calculation:

  $\begin{array}{c}P\left(B\right)=\frac{300}{595}\\ =0.5042\\ P\left(M\right)=\frac{320}{595}\\ =0.5378\end{array}$

There

  $P\left(B\right)\ne P\left(B|M\right)andP\left(M\right)\ne P\left(M|B\right)$ it is indicating that the events D and F not independent.