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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.

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Best answer

(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.

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