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Exercise 12.3 · Q3

Q.If AA and BB are two independent events such that P(A∪B)=0.6P(A\cup B) = 0.6, P(A)=0.2P(A) = 0.2, find P(B)P(B).

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✓ Free question

Step 1. Independent-event Addition Theorem. Since A,BA,B are independent, P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B), so P(A∪B)=P(A)+P(B)−P(A)P(B)P(A\cup B)=P(A)+P(B)-P(A)P(B).

Step 2. Substitute known values. 0.6=0.2+P(B)−(0.2)P(B)=0.2+0.8 P(B)0.6=0.2+P(B)-(0.2)P(B)=0.2+0.8\,P(B).

Step 3. Solve for P(B)P(B). 0.8 P(B)=0.6−0.2=0.4 ⇒ P(B)=0.40.8=0.50.8\,P(B)=0.6-0.2=0.4\ \Rightarrow\ P(B)=\dfrac{0.4}{0.8}=0.5.

✓Final answer

P(B)=0.5P(B)=0.5.

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