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

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)

∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74

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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).

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) ∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74

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)

∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74