Q.

A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3.The probability of B getting selected cannot exceed.

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answer is 0000.80.

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Detailed Solution

Given P(A)=0.5 and P(A∩B)≤0.3  we have P(A∪B)=P(A)+P(B)-P(A∩B)≤1⇒P(A)+P(B)≤1+P(A∩B)≤1+0.3=1.3⇒0.5+P(B)≤1.3⇒P(B)≤0.8Thus probability of B getting selected cannot exceed 0.8
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