1. ## Mixed Probability

Events A & B are such that P(AB) = 1/3, P(B|A)= 1/4, and P(B'|A') = 4/5.

Hence find a) P(B'|A) b)P(AnB) c) P(B) d) P(AUB)

I've been staring at it for 1/2 hour and still haven't got anywhere

2. Do you know that $\displaystyle P(B|A) = \frac{P(A\cap B)}{P(A)}$ ?

3. P(1/4) = P(AnB)/P(1/2)

So P(AnB) = 1/8 ( I think)

I'm slowly getting there...

4. Originally Posted by Wildsurf
P(1/4) = P(AnB)/P(1/2)

So P(AnB) = 1/8 ( I think)
No indeed. In fact part of the given is $P(AB)=P(A\cap B)=\frac{1}{3}$

5. P(AnB) = P(A)P(B)

1/8 = (1/3)P(B)

1/8 = P(B)
----
1/3

1/8 X 3/1 = P(B)

3/8 = P(B)

6. Originally Posted by Plato
No indeed. In fact part of the given is $P(AB)=P(A\cap B)=\frac{1}{3}$
My bad P(AB) was in fact meant to be P(A)

7. For a) $P(B^c|A)=1-P(B|A)$.
BUT $P(B|A^c)\ne 1-P(B|A)$.