Events $A$ and $B$ are mutually exclusive. Suppose event $A$ occurs with probability $0.74$ and event $B$ occurs with probability $0.22$.
- Compute the probability that $A$ occurs but $B$ does not occur.
- Compute the probability that either $A$ occurs without $B$ occurring or $A$ and $B$ both occur.
Answers:
- $\Pr(A \cap \neg B) = \Pr(A) + \Pr(\neg B)- \Pr(A \cap B) = 0.74 + 0.78 -(0.74 \cdot 0.78)=0.9428$
- $\Pr([B \cap A] \cup [A \cap B])= \Pr(\neg B \cap A) + \Pr(A \cap B) = 0.78 \cdot 0.74 + 0.74 \cdot 0.22=0.74$
Is it correct?
Best Answer
Supposing, as is usual (e.g. here) that mutually exclusive means that $P(A \cap B) = 0$ we have that $P(A \land \lnot B) = P(A \setminus B) = P(A) - P(A \cap B) = P(A) = 0.74$.
The final one is then just $P(A)$ too. $A$ cannot occur at the same time as $B$.