Question

Let A and B be two events in a sample space S such that P(A) = 0.5, P(B) = 0.6, and P(A intersectionB) = 0.15. Find the probabilities below. Hint: (A intersectionBc) union (A intersectionB) = A.

(a) P(A|Bc)

(b) P(B|Ac)

Answer 1

(a)
`(7)/(8)`

(b)
`(9)/(10)`

Explanation:

Given,

P(A) = 0.5 ⇒
`P(A^c)=1-P(A) = 1 - 0.5 = 0.5`

P(B) = 0.6 ⇒
`P(B^c)=1-P(B) = 1 - 0.6 = 0.4`

P(A∩B) = 0.15

∵
`P(A\cap B^c)=P(A) - P(A\cap B) = 0.5 - 0.15 = 0.35`

Similarly,

`P(B\cap A^c)=P(B) - P(B\cap A) = 0.6 - 0.15 = 0.45`

Now,

(a)
`P((A)/(B^c))=(P(A\cap B^c))/(P(B^c))=(0.35)/(0.4)=(35)/(40)=(7)/(8)`

(b)
`P((B)/(A^c))=(P(B\cap A^c))/(P(A^c))=(0.45)/(0.5)=(45)/(50)=(9)/(10)`