# Probability that event $A$ occurs but event $B$ does not occur when events $A$ and $B$ are mutually exclusive

Events $A$ and $B$ are mutually exclusive. Suppose event $A$ occurs with probability $0.74$ and event $B$ occurs with probability $0.22$.

1. Compute the probability that $A$ occurs but $B$ does not occur.
2. Compute the probability that either $A$ occurs without $B$ occurring or $A$ and $B$ both occur.

1. $\Pr(A \cap \neg B) = \Pr(A) + \Pr(\neg B)- \Pr(A \cap B) = 0.74 + 0.78 -(0.74 \cdot 0.78)=0.9428$
2. $\Pr([B \cap A] \cup [A \cap B])= \Pr(\neg B \cap A) + \Pr(A \cap B) = 0.78 \cdot 0.74 + 0.74 \cdot 0.22=0.74$

Is it correct?

• What is your definition of mutually exclusive events? Apr 6, 2018 at 9:21
• That means very different from each other? Apr 6, 2018 at 9:22
• That's not a mathematical definition. One definition is $P(A \cap B) = 0$. Apr 6, 2018 at 9:23
• ohhhh!!!!!! so for no.1 it should be 0.74+0.78? Apr 6, 2018 at 9:25
• Can a probability be $>1$? Apr 6, 2018 at 9:25

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 $B$.
• Mutually exclusive implies that $P(A\cap B)=0$ which is enough here. The actual meaning of it is that $A\cap B=\varnothing$. Apr 6, 2018 at 9:58
• @drhab Quite. And so both parts are just roundabout ways of asking for $P(A)$. A bit confusing for the OP maybe. Apr 6, 2018 at 10:48