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Exercise 12.2 · Q4

Q.The probability of an event AA occurring is 0.5 and BB occurring is 0.3. If AA and BB are mutually exclusive events, then find the probability of

(i) P(A∪B)P(A\cup B)
(ii) P(A∩Bˉ)P(A\cap \bar B)
(iii) P(Aˉ∩B)P(\bar A\cap B).
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Step 1. Given. P(A)=0.5P(A)=0.5, P(B)=0.3P(B)=0.3, A,BA,B mutually exclusive (P(A∩B)=0P(A\cap B)=0).

Step 2. Part (i). P(A∪B)=P(A)+P(B)=0.5+0.3=0.8P(A\cup B)=P(A)+P(B)=0.5+0.3=0.8.

Step 3. Part (ii). Since A,BA,B are disjoint, AA lies entirely inside Bˉ\bar B, so A∩Bˉ=AA\cap\bar B=A: P(A∩Bˉ)=P(A)−P(A∩B)=0.5−0=0.5P(A\cap\bar B)=P(A)-P(A\cap B)=0.5-0=0.5. …

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