Event A and B are such that P(A) = 0.5 , P(B) = 0.7,P(A ∩ B) = 0.2,find P(A ∩ B’).
Answers
Answer:
1. Assuming A, B, C are mutually independent, with P(A) = P(B) = P(C) = 0.1,
compute:
(a) P(A ∪ B) Solution: P(A) + P(B) − P(A)P(B) = 0.19
(b) P(A ∪ B ∪ C)
Solution: By formula the formula for P(A∪B∪C) and indep., P(A∪B∪C) =
3 · 0.1 − 3 · 0.1
2 + 0.1
3 = 0.271
(c) P(A \ (B ∪ C))
Solution: P(A) − P(A ∩ B) − P(A ∩ C) + P(A ∩ B ∩ C) = 0.081
2. Given that P(A) = 0.3, P(A|B) = 0.4, and P(B) = 0.5, compute:
(a) P(A ∩ B) Solution: P(A|B)P(B) = 0.4 · 0.5 = 0.2
(b) P(B|A) Solution: P(B ∩ A)/P(A) = 0.2/0.3 = 0.666
(c) P(A0
|B) Solution: P(A0 ∩ B)/P(B) = ((P(B) − P(A ∩ B))/P(B) = 0.6
(d) P(A|B0
) Solution: P(A∩B0
)/P(B0
) = (P(A)−P(A∩B))/(1−P(B)) = 0.2
3. Assume A and B are independent events with P(A) = 0.2 and P(B) = 0.3. Let C be
the event that at least one of A or B occurs, and let D be the event that exactly
one of A or B occurs.
(a) Find P(C).
Solution: The event C is just the union of A and B, so P(C) = P(A ∪ B) =
P(A) + P(B) − P(A)P(B) = 0.44