Question

A and B are two candidates seeking admission to IIT. The probability that A getting selecte
is 0.5 and the probability that both A and B getting selected is 0.3. Prove that the
probability of B being selected is at most 0.8​

Answer 1

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.