A pair of fair dice is tossed. Define the following events:

A: [Exactly one of the dice shows a 1.]

B: [The sum of the numbers on the two dice is even.]

a. Identify the sample points in the events $\mathbf{A}\mathbf{,}\mathbf{B}\mathbf{,}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{,}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{,}{\mathbf{andA}}^{\mathbf{c}}\mathbf{.}$

b. Find the probabilities of all the events from part a by summing the probabilities of the appropriate sample points.

C. Using your result from part b, explain why A and B are not mutually exclusive.

d. Find $\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}$ using the additive rule. Is your answer the same as in part b?

The sample points of the sample space are the primary outcomes of an experiment. Sample points are sample space elements that represent the experiment in terms of the sample space. Sample points are often referred to as sampling units or observations.

Rolling a pair of fair dice gives the following sample points:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

$\mathbf{A}\mathbf{,}\mathbf{B}\mathbf{,}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{,}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{,}{\mathbf{andA}}^{\mathbf{c}}\mathbf{.}$

$\mathbf{A}\mathbf{=}\mathbf{\left[}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\right]}$

$\begin{array}{l}\mathbf{B}\mathbf{=}\mathbf{\left[}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{\right(}\mathbf{4}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{]}\end{array}$

$\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{=}\mathbf{\left[}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{\right]}$

$\begin{array}{l}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{=}\mathbf{[}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{1}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{]}\end{array}$

$\begin{array}{l}{\mathbf{A}}^{\mathbf{c}}\mathbf{=}\mathbf{\left[}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{\right(}\mathbf{4}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{6}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{5}\mathbf{)}\mathbf{(}\mathbf{6}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{]}\end{array}$

Here, the total number of sample points is 36.

Likewise,

$\begin{array}{l}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{36}\\ \mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{11}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{n}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\mathbf{18}\\ \mathbf{n}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{5}\end{array}$

$\begin{array}{c}\mathbf{n}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{24}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{n}\mathbf{\left(}{\mathbf{A}}^{\mathbf{c}}\mathbf{\right)}\mathbf{=}\mathbf{25}\end{array}$

Now, summing the probabilities of the data from a sample point is:

$\begin{array}{c}\mathbf{}\mathbf{}\mathbf{}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{11}}{\mathbf{36}}\mathbf{,}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\frac{\mathbf{18}}{\mathbf{36}}\mathbf{,}\\ \mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{36}}\mathbf{,}\\ \mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{24}}{\mathbf{36}}\mathbf{,}\end{array}$

$\mathbf{P}\mathbf{\left(}{\mathbf{A}}^{\mathbf{c}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{36}}$

If$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{0}$, A and B are hence mutually exclusive.

Here,

$\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{5}}{\mathbf{36}}\mathbf{\ne }\mathbf{0}$

Hence, A and B don't get to be mutually exclusive.

$\begin{array}{l}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{11}}{\mathbf{36}}\mathbf{+}\frac{\mathbf{18}}{\mathbf{36}}\mathbf{-}\frac{\mathbf{5}}{\mathbf{36}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{11}\mathbf{+}\mathbf{18}\mathbf{-}\mathbf{5}}{\mathbf{36}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{24}}{\mathbf{36}}\end{array}$

Yes, it is the same as in part b.