Chapter 7.2, Problem 7.37E

(a)

To determine

To construct: a two-way table that displays the sample space in terms of events B and E.

Explanation of Solution

Displaying the two-way table which have the sample space in terms of events B and E

	B	Not B	Total
E	9	11	20
Not E	9	9	18
Total	18	20	38

(b)

To determine

To Calculate: the $\text{P}\left(\text{B}\right)\text{and}\text{P}\left(\text{E}\right)$ .

Answer to Problem 7.37E

  $\begin{array}{c}\text{P}\left(\text{B}\right)=0.4737\\ \text{P}\left(\text{E}\right)=0.5263\end{array}$

Explanation of Solution

Given:

	B	Not B	Total
E	9	11	20
Not E	9	9	18
Total	18	20	38

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

  $\begin{array}{c}\text{P}\left(\text{B}\right)=\frac{18}{38}\\ =0.4737\\ \text{P}\left(\text{E}\right)=\frac{20}{38}\\ =0.5263\end{array}$

(c)

To determine

To Explain: the event “B and E” in word and find $\text{P}\left(\text{B and E}\right)$ .

Answer to Problem 7.37E

  $\text{P}\left(\text{B and E}\right)=0.2368$

Explanation of Solution

Given:

	B	Not B	Total
E	9	11	20
Not E	9	9	18
Total	18	20	38

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

“B and E” is a phenomenon that moves the ball to a right angled slot.

  $\begin{array}{c}\text{P}\left(\text{B and E}\right)=\frac{9}{38}\\ =0.2368\end{array}$

(d)

To determine

To Calculate: the “B or E” in words and find the $\text{P}\left(\text{B or E}\right)$

Answer to Problem 7.37E

  $\text{P}\left(\text{B or E}\right)=0.7632$

Explanation of Solution

Given:

	B	Not B	Total
E	9	11	20
Not E	9	9	18
Total	18	20	38

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

"B and E" is the event that the ball lands on an even or a black slot.

  $\begin{array}{c}\text{P}\left(\text{B or E}\right)=\frac{9+11+9}{38}\\ =\frac{29}{38}\\ =0.7632\end{array}$