An experiment results in one of three mutually exclusive events, A, B, or C. It is known that P (A)= .30, P(B)= .55 , and P(C)= .15. Find each of the following probabilities:

a. $\mathbf{\text{P(A}}\mathbf{\cup }\mathbf{\text{B)}}$

b.$\mathbf{\text{P(A}}\mathbf{\cap }\mathbf{\text{C)}}$

c. P (A/B)

d. $\mathbf{\text{P(B}}\mathbf{\cup }\mathbf{\text{C)}}$

e. Are B and C independent events? Explain.

When two events have no components (their intersection is the empty set), they are said to be mutually exclusive. Therefore, $\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B) = 0}}$. It signifies that the chances of events A and B occurring are nil.

Since A and B are mutually exclusive.

Therefore,

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

By addition theorem on probability:

$\begin{array}{c}\mathbf{\text{P(A}}\mathbf{\cup }\mathbf{\text{B) = P(A) + P(B)}}\mathbf{-}\mathbf{\text{P(A}}\mathbf{\cap }\mathbf{\text{B)}}\\ \mathbf{\text{= 0.30 + 0.55}}\mathbf{-}\mathbf{\text{0}}\\ \mathbf{\text{= 0.85}}\end{array}$

Hence, the required probability is 0.85.

Since A and C are mutually exclusive.

Therefore,

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

Hence, the required probability is 0.

$\begin{array}{c}\mathbf{\text{P (A/B)=}}\frac{\mathbf{\text{P (A}}\mathbf{\cap }\mathbf{\text{B)}}}{\mathbf{\text{P (B)}}}\\ \mathbf{\text{=0}}\end{array}$

Hence, the required probability is 0.

$\begin{array}{c}\mathbf{\text{P (B}}\mathbf{\cup }\mathbf{\text{C) = P (B) + P (C)}}\mathbf{-}\mathbf{\text{P (B}}\mathbf{\cap }\mathbf{\text{C)}}\\ \mathbf{\text{= 0.55 + 0.15}}\mathbf{-}\mathbf{\text{0}}\\ \mathbf{\text{= 0.70}}\end{array}$

Hence, the required probability is 0.70.

Yes, the occurrences are independent because the occurrence of one guarantees that the others will not occur. As a result, B and C are independent.