Let A, B, and C be events such that A and B are independent, B and C are mutually exclusive, P[A]=1/4, P[B]=1/6 and P[C]=1/2. What is P[(A ∩ B)' U C]? Answer 23/24. How did they get that?
2006-11-20 21:12:47 · 2 answers · asked by tkbrooks1980 1 in Science & Mathematics ➔ Mathematics
P(A ∩ B) = 1/4 * 1/6 = 1/24, since they are independent. So P((A ∩ B)') = 23/24. But what about the C? Well, that doesn't actually matter at all, because B and C are mutually exclusive. If you draw a Venn diagram, you'll notice that all of C is already accounted for when you've coloured in (A ∩ B)'.
2006-11-20 21:19:31 · answer #1 · answered by stephen m 4 · 0⤊ 0⤋
Did you not make a mistake in your answer 23/24
The probability of a intersect p is P(A) * P(B) = 1/4*1/6 = 1/24
The probability of C is P (C) =1/2 = 12/24
Since the b anc is exclusive
the probability of union is 12/24+1/24 = 13/24
For the answers I have not the sign unionon my board
2006-11-21 05:25:59 · answer #2 · answered by maussy 7 · 0⤊ 0⤋