Question

A and B appear for an interview for two vacancies in the same post. The probability of A’s selection is 1/6 and that of B’s selection is 1/4. Find the probability that (i) both of them are selected (ii) only one of them is selected (iii) none is selected (iv) at least one of them is selected. Select the correct answer from above options

Answer 1

(i) Probability that both of them are selected P (both of them are selected) = P (A ∩ B) = P (A) × P (B) Substituting the values = 1/6 × 1/4 = 1/24 Hence, the probability that both of them are selected is 1/24. (ii) Probability that only one of them is selected P (only one of them is selected) = P (A and not B or B and not A) = P (A and not B) + P (B and not A) We get Substituting the values = (1/6 × 3/4) + (1/4 × 5/6) = 3/24 + 5/24 So we get = 1/3 Hence, the probability that only one of them is selected is 1/3. (iii) Probability that none is selected P (none is selected) Substituting the values = 5/6 × 3/4 = 5/8 Hence, the probability that none is selected is 5/8. (iv) Probability that at least one of them is selected P (at least one of them is selected) = P (selecting only A) + P (selecting only B) + P (selecting both) We know that = P (A and not B) + P (B and not A) + P (A and B) It can be written as Substituting the values = (1/6 × 3/4) + (1/4 × 5/6) + (1/6 × 1/4) On further calculation = 3/24 + 5/24 + 1/24 = 3/8 Hence, the probability that at least one of them is selected in 3/8.