If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find (i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

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Solution

The given probabilities are,

P( A )=0.8, P( B )=0.5 and P( B|A )=0.4

(i)

The probability P( A∩B ) is calculated as,

P( A∩B )=P( B|A )×P( A ) =0.4( 0.8 ) =0.32

Therefore, the value of P( A∩B ) is 0.32.

(ii)

The Probability P( A|B ) is calculated as,

P( A|B )= P( A∩B ) P( B ) = 0.32 0.5 =0.64

Therefore, the value of P( A|B ) is 0.64.

(iii)

The Probability P( A∪B ) is calculated as,

P( A∪B )=P( A )+P( B )−P( A∩B ) =0.8+0.5−0.32 =1.3−0.32 =0.98

Therefore, the value of P( A∪B ) is 0.98.

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