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A and B are two candidates seeking admission to IIT. The probability that A getting selecteis 0.5 and the probability that both A and B getting selected is 0.3. Prove that theprobability of B being selected is at most 0.8 |
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Answer» P(A) = 0.5 P(A ∩ B) ≤ 0.3 Now, USING the relationship between the PROBABILITY of two events ⇒ P(A) + P(B) - P(A ∩ B) = P(A ∪ B) ⇒ 0.5 + P(B) - P(A ∩ B) = 1 ⇒ P(B) - P(A ∩ B) = 0.5 ⇒ P(B) = 0.5 + P(A ∩ B) ⇒ P(B) ≤ 0.5 + 0.3 ⇒ P(B) ≤ 0.8 Thus, The Probability of B being selected is at most 0.8 Hence PROVED. |
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