Question:
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)
(iii) P(A or B).
Solution:
It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16
(i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) We know that P(A or B) = P(A) + P(B) – P(A and B)
$\therefore P(A$ or $B)=0.42+0.48-0.16=0.74$
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