Chapter 7.2, Problem 7.34E

(a)

To determine

To construct: a two-way table that describes the sample space in terms of events J and R.

Explanation of Solution

J = getting a jack and event R= getting a red card.

Sample space in terms of events J and R

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

(b)

To determine

To Calculate: $\text{P}\left(\text{J}\right)\text{,P}\left(\text{R}\right)\text{and}\text{P}\left(\text{J}\text{and}\text{R}\right)$

Answer to Problem 7.34E

  $\begin{array}{c}\text{P}\left(\text{J}\right)=0.0769\\ \text{P}\left(\text{R}\right)=0.50\\ \text{P}\left(\text{J}\text{and}\text{R}\right)=0.0385\end{array}$

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

  $\begin{array}{c}\text{P}\left(\text{J}\right)=\frac{4}{52}\\ =\frac{1}{13}\\ =0.0769\\ \text{P}\left(\text{R}\right)=\frac{26}{52}\\ =\frac{1}{2}\\ =0.50\end{array}$

  $\begin{array}{c}\text{P}\left(\text{J}\text{and}\text{R}\right)=\frac{2}{52}\\ =0.0385\end{array}$

(c)

To determine

To Explain: that why $\text{P}\left(\text{J}\text{or}\text{R}\right)\ne \text{P}\left(\text{J}\right)+\text{P}\left(\text{R}\right)$ and use the general addition rule to compute $\text{P}\left(\text{J}\text{or}\text{R}\right)$ .

Answer to Problem 7.34E

  $\text{P}\left(\text{J}\text{or}\text{R}\right)=0.5384$

Explanation of Solution

Formula used:

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

Calculation:

Because $\text{P}\left(\text{J}\text{and}\text{R}\right)$ is not equal to zero that’s why $\text{P}\left(\text{J}\text{or}\text{R}\right)\ne \text{P}\left(\text{J}\right)+\text{P}\left(\text{R}\right)$ so using the addition rule

  $\begin{array}{c}\text{P}\left(\text{J}\text{or}\text{R}\right)=\text{P}\left(\text{J}\right)+\text{P}\left(\text{R}\right)-\text{P}\left(\text{J}\text{and}\text{R}\right)\\ =0.0769+0.50-0.0385\\ =0.5384\end{array}$