we know ,P(B | A)+ P(B’|A) = 1

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.

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.