# A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16.

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 $B)=0.42+0.48-0.16=0.74$