Question on probability: Concerning the events of A and B?

Question

P(B)=1/4 P(A|B)=1/2 P(A|B=1/3 (there is a straight line over the B.

Determine (a) P(AB) there is a straight line above the B
(b) P(A) (c) P(B|A) there is a straight line above the B

Answer 1

Given

P(B) = 1/4
P(A|B) = 1/2
P(A|B bar) =1/3

I assume that by B bar you mean the complement of B.

P(B) + P(B bar) = 1
P(B bar) = 1 - P(B) = 1 - 1/4 = 3/4

(a) Find P(A ∩ B bar)
P(A ∩ B bar) = P(B bar) P(A | B bar) = (3/4)(1/3) = 1/4

P(A ∩ B) = P(B) P(A | B ) = (1/4)(1/2) = 1/8

(b) Find P(A).
P(A) = P(A ∩ B) + P(A ∩ B bar) = 1/4 + 1/8 = 3/8

(c) Find P(B bar | A)
Find P(B bar | A) = P(A ∩ B bar) / P(A) = (1/4) / (3/8) = 2/3

Answer 2

Remember the formula
P(X|Y) = P(XY) / P(Y)
or P(X|Y)*P(Y) = P(XY)
I will write ~B instead of B with a bar over it.
So we are given
P(B) = 1/4, P(A|B) = 1/2, and P(A|~B) = 1/3.

Since P(B) = 1/4, we get P(~B) = 1 - P(B) = 3/4

(a) P(A~B) = P(A|~B)*P(~B) = (1/3)*(3/4) = 1/4

(b) P(AB) = P(A|B)*P(B) = (1/2)*(1/4) = 1/8
Then P(A) = P(AB) + P(A~B) = 1/8 + 1/4 = 3/8

(c) P(B|A) = P(AB) / P(A) = (1/8) / (3/8) = 1/3

Answer 3

tough