A sample space contains six sample points and events A, B, and C as shown in the Venn diagram. The probabilities of the sample points are

$\begin{array}{l}\mathbf{\text{P (1) = .20, P (2) = .05, P (3) = .30, P (4) = .10,}}\\ \mathbf{\text{P (5) = .10, P (6) = .25.}}\end{array}$

a. Which pairs of events, if any, are mutually exclusive? Why?

b. Which pairs of events, if any, are independent? Why?

c. Find$\mathbf{\text{P (A}}\mathbf{\cup }\mathbf{\text{B)}}$ by adding the probability of the sample points and then using the additive rule. Verify that the answers agree. Repeat for$\mathbf{\text{P (A}}\mathbf{\cup }\mathit{C}\mathbf{\text{)}}$

The relationship between two or more occurrences that cannot occur simultaneously is referred to as mutually exclusive. If A and B cannot exist simultaneously, they are mutually exclusive. They suggests that A and B have no common outcomes, and$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0}}$

Here, we have

$\begin{array}{c}\mathbf{\text{P (A) = [1, 2, 3] = [.20 + .05 + .30] = .55}}\\ \mathbf{\text{P (B) = [3, 4] = [.30 + .10] = .40}}\\ \mathbf{\text{P (C) = [5, 6] = [.10 + .25] = .35}}\\ \mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = [3] = .30}}\end{array}$

For the two events to be mutually exclusive, only one of the two occurrences may occur at the same moment, indicating that:

$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0}}$

Here, the pairs are

$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B),P (A}}\mathbf{\cap }\mathit{C}\mathbf{\text{),P (B}}\mathbf{\cap }\mathit{C}\mathbf{\text{)}}$

Since,$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0.30P (A}}\mathbf{\cap }\mathbf{\text{C) = 0,P (B}}\mathbf{\cap }\mathbf{\text{C) = 0}}$

Therefore, the two pairs of events that are mutually exclusive are A and C & B and C.

To be independent, the likelihood of one event occurring does not affect the probability of the other, indicating that:

$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = P (A) x P (B)}}$

Here, the pairs are:

$\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B),P (A}}\mathbf{\cap }\mathit{C}\mathbf{\text{),P (B}}\mathbf{\cap }\mathit{C}\mathbf{\text{)}}$

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0.55}}\mathbf{\times }\mathbf{\text{0.40}}\\ \mathbf{\text{= 0.22}}\end{array}$

$\begin{array}{c}\mathbf{\text{P (B}}\mathbf{\cap }\mathbf{\text{C) = 0.40}}\mathbf{\times }\mathbf{\text{0.35}}\\ \mathbf{\text{= 0.14}}\end{array}$

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{C) = 0.55}}\mathbf{\times }\mathbf{\text{0.35}}\\ \mathbf{\text{= 0.1925}}\end{array}$

Therefore,

$$\begin{gathered} \mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0.22}}\mathbf{\ne }\mathbf{\text{0}} \\ \mathbf{\text{P (B}}\mathbf{\cap }\mathbf{\text{C) = 0.14}}\mathbf{\ne }\mathbf{\text{0}} \\ \mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{C) = 0.65}}\mathbf{\ne }\mathbf{\text{0}} \end{gathered}$$

Hence, Nopair is independent.

We know that the addictive rule to find the probability is:

$\mathbf{\text{P (A}}\mathbf{\cup }\mathbf{\text{B) = P (A) + P (B) - P(A}}\mathbf{\cap }\mathbf{\text{B)}}$

L.H.S

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{\cup }\mathbf{\text{B) = [1, 2, 3, 4]}}\\ \mathbf{\text{= [.20 + .05 + .30 + .10]}}\\ \mathbf{\text{= .65}}\end{array}$

R.H.S

$\begin{array}{c}\mathbf{\text{P (A) + P (B)}}\mathbf{-}\mathbf{\text{P(A}}\mathbf{\cap }\mathbf{\text{B)}}\\ \mathbf{\text{=.55 + .40}}\mathbf{-}\mathbf{\text{.30}}\\ \mathbf{\text{=.65}}\end{array}$

L.H.S = R.H.S

Same with$\mathbf{\text{P (A}}\mathbf{\cup }\mathbf{\text{C)}}$

L.H.S

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{\cup }\mathbf{\text{C) = [1, 2, 3, 5, 6]}}\\ \mathbf{\text{= [.20 + .05 + .30 + .10 + .25]}}\\ \mathbf{\text{= .90}}\end{array}$

R.H.S

$\begin{array}{c}\mathbf{\text{P (A) + P (C)}}\mathbf{-}\mathbf{\text{P(A}}\mathbf{\cap }\mathbf{\text{C)}}\\ \mathbf{\text{=.55 + .35}}\mathbf{-}\mathbf{\text{0}}\\ \mathbf{\text{=.90}}\end{array}$

L.H.S = R.H.S

Yes, the answer agrees in both cases.