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Question

# (a) If A and B be mutualy exclusive events associated with a random experiment such that P(A) = 0.4 and P(B) = 0.5, then find : (i) P(A∪B) (ii) P(¯¯¯¯A∪¯¯¯¯B) (iii) P(¯¯¯¯A∪B) (iv) P(A∪¯¯¯¯B) (b) A and B are two events such that P(A)= 0.54, P(B) = 0.69 and P(A∪B) = 0.35. Find: (i) P(A∪B) (ii) P(¯¯¯¯A∪¯¯¯¯B) (iii) P(A∪¯¯¯¯B) (iv) P(B∪¯¯¯¯A) (c) Fill in the blanks in the following table : P(A) P(B) P(A∩B) P(A∪B) (i) 13 15 115___ (ii) 0.35 ___ 0.25 0.6 (iii) 0.5 0.35 ___ 0.7

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Solution

## (a) Given, P(A) = 0.4 P(B) = 0.5 ∴ A and B are mutually exclusive events, then P(A∩B)=0 Now, (i) P(A∪B)=p(A)+P(B)−P(A∩B) =0.4+0.5 - 0 =0.9 ∴P(A∪B)=0.9 (ii) P(¯¯¯¯A∩¯¯¯¯B)=1−P(A∪B) = 1-0.9 = 0.1 ∴P(¯¯¯¯A∩¯¯¯¯B)=0.1 (iii) P(¯¯¯¯A∩B)=P(B)−P(A∩B) = 0.5-0 ∴P(¯¯¯¯A∩B)=0.5 (iv) P(A∩¯¯¯¯B)=P(A)−P(A∩B) = 0.4-0 = 0.4 ∴P(¯¯¯¯A∩B)=0.4 (b) Given, P(A) = 0.54 P(B) = 0.69 P(A∩B)=0.35 (i) P(A∪B)=P(A)+P(B)−P(A∩B) = 0.54 + 0.69 -0.35 = 1.23 - 0.35 ∴P(A∪B)=0.88 (ii) P(¯¯¯¯A∩¯¯¯¯B)=1−P(A∪B) = 1-0.88 = 0.12 ∴P(¯¯¯¯A∩¯¯¯¯B)=0.12 (iii) P(A∩¯¯¯¯B)=1−P(A∪B) = 0.54-0.35= 0.19 ∴P(A∩¯¯¯¯B)=0.19 (iv) P(B∩¯¯¯¯A)=P(B)−P(A∩B) = 0.69 - 0.35 = 0.34 ∴P(B∩¯¯¯¯A)=0.34 (c) (i) Given, P(A)=13,P(A∩B)=115 P(B)=15,P(A∪B) = ∵P(A∪B)=P(A)+P(B)−P(A∩B) = 13+15−115 = 5+3−115=8−115=715 ∴P(A∪B)=715 (ii) Given, P(A) = 0.35, P(B)= ___ P(A∩B)=0.25,P(A∪B)=0.6 ∵P(A∪B)=P(A)+P(B)=P(A∩B) 0.6 = 0.35+ P(B) - 0.25 0.6 = 0.10 + P(B) P(B) = 0.6 -0.1 P(B) = 0.5 (iii) Given, P(A) = 0.5, P(B) = 0.35 P(A∩B) = ___ P(A∪B)=0.7 ∵P(A∪B)=P(A)+P(B)−P(A∩B) 0.7 = 0.5 +0.35 - P(A∩B) 0.7 = 0.85 - P(A∩B) P(A∩B) = 0.85 - 0.7 P(A∩B) = 0.15

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