Question

Let A and B be two events in a sample space S such that P(A) = 0.6, P(B) = 0.5, and P(A B) = 0.15. Find the probabilities P(A|B^c) -->(B compliment)

Answer 1

Can someone give me the generic formula for this too? It always confuses me.

Answer 2

you can most easily do it with a venn diagram rather than a formula, but
\[P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)}\]

Answer 3

So would the answer just ne p(A)?

Answer 4

since \[P(B)=.5\] you know
\[P(B^c)=1-.5=.5\]

Answer 5

(.6 * .5)/.5?

Answer 6

no no hold on

Answer 7

you cannot multiply the probabilities together to get the intersection unless they are independent
\[P(A\cap B^c)=.45\]

Answer 8

let me draw a picture to make it clear

Answer 9

since the probability of the intersection of A and B is .15, and since the probability of A is .6 you get the probability of \(A\cap B^c =.45\)

Answer 10

so your "final answer" is
\[\frac{.45}{.5}=\frac{45}{50}=.9\]

Answer 11

there was a mistake in my picture, the number outside should have been 5