given,

P(B’|A) P(A) = p3

P(B|A) P(A) = p2...(1)

P(B|A) = 1-(p3/P(A))

So, P(B|A) P(A) = P(A) -p3……...(2)

and from 1 and 2

P(A) = p2+p3.

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Well, from venn diagram it looks pretty obvious that the

P(A) = P1 + P2

But, I tried another approach...

@Arjun sir, can you please check this out..

P(A) = P(A/B) + P(A/B')

That is, the probability of occurrence of A is the sum of:

i) Probability that A occurs given that B has occurred: P(A/B)

ii) Probability that A occurs given that B' has occurred: P(A/B')

Since, B and B' totally make up sample space, the occurrence of A is also covered.

P(A/B) = __P(A∩B) __ = __P2__

P(B) P1

P(A/B') = __P(A∩B') __ = __ P3 __

P(B') 1-P1

P(A) = __P2 __ + __ ____P3 __

P1 1-P1