can you check my answers for this question?
for c i got 0.3 * 0.5 = 0.15
for b i multiplied the outcome of a by b compliment, but b compliment is still .5, so is the answer the same as c?
and for a i know it means a union b but i dont know how to calculate it?
Suppose that A and B are mutually exclusive events for which
P(A) = 0.3 and P(B) = 0.5. What is the probability that
(a) either A or B occurs?
(b) A occurs but B does not?
(c) both A and B occur?
Best Answer
If A and B are mutually exclusive then $P(A\cap B) = 0$.
For a), take into account that $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. For b), note that the probability of $B$ not ocurring is $P(B^c)$, and c) should be clear.
Edit:
$A\cup B = (A\setminus B)\cup (A\cap B) \cup (B\setminus A)$, where the sets are disjoint. It is a principle that the probability of the union of disjoint sets is the sum of the probabilities of each set of the union. In particular $$P(A\cup B) = P(A\setminus B)+ P(A\cap B) +P(B\setminus A).$$
Also $A\setminus B = A \setminus (A\cap B)$. Note that $[A \setminus (A\cap B)]\cup [A\cap B] = A.$ Since the sets in brackets are disjoint: $P(A\setminus B) + P(A\cap B) = P(A)$. Symmetrically $P(B) = P(B\setminus A) + P(A\cap B)$. It then follows that $P(A\cup B) = P(A) + P(B) - P(A\cap B)$.