A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P (not B) and (iii) P(A or B).

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Solution

Given, A and B are two events such that,

P( A )=0.42 P( B )=0.48 P( A and B )=0.16

(i)

The probability P( not A ) is,

P( notA )=1−P( A ) =1−0.42 =0.58

Thus, P( not A )is 0.58.

(ii)

The probability P( not B ) is,

P( not B )=1−P( B ) =1−0.48 =0.52

Thus, P( not B ) is 0.52.

(iii)

The probability P( A or B ) is,

P( A or B )=P( A )+P( B )−P( A and B ) =0.42+0.48−0.16 =0.74

Thus, P( A or B )is 0.74.

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