Chapter 5.2, Problem 52E

a)

To determine

To make a two-way table that displays the sample space in terms of events J and R.

Explanation of Solution

Given:

The events: getting jack and getting red card.

The two-way table is,

	J	Not J	Total
R	2	24	26
Not R	2	24	26
Total	4	48	52

b)

To determine

To find P(J) and P(R)

Answer to Problem 52E

P(J) = 0.077 and P(R)= 0.5

Explanation of Solution

Given:

	J	Not J	Total
R	2	24	26
Not R	2	24	26
Total	4	48	52

Formula:

  $Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

P(jack) = P(J)

  $=\frac{4}{52}=\frac{1}{13}$

P(Red)=P(R)

  $=\frac{26}{52}=\frac{1}{2}$

c)

To determine

To find P(J and R)

Answer to Problem 52E

P(J and R) = $\frac{1}{26}$

Explanation of Solution

Given:

	J	Not J	Total
R	2	24	26
Not R	2	24	26
Total	4	48	52

Formula:

  $Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

P(J and R)

  $=\frac{2}{52}=\frac{1}{26}$

d)

To determine

To find P(J or R)

Answer to Problem 52E

P(J or R) = $\frac{7}{13}$

Explanation of Solution

Given:

P(J and R) = $\frac{1}{26}$

P(J) = 0.077 and P(R)= 0.5

Formula:

  $Probability=\frac{Numberoffavorableoutcomes}{Totalpossibleoutcomes}$

  $P\left(\text{J }or\text{ R}\right)=P\left(J\right)+P\left(R\right)-P\left(\text{J }and\text{ R}\right)$

As J and R are not mutually exclusive,

  $P\left(\text{J }or\text{ R}\right)\ne P(J)+P(R)$

That is J and R occurs together.

Therefore,

  $P\left(\text{J }or\text{ R}\right)=\frac{1}{13}+\frac{1}{2}-\frac{1}{26}=\frac{7}{13}$