# 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.

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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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Given : A and B appear for an interview ,then P(A) = $\frac{1}{6}$ and P(B) = $\frac{1}{4}$

$\Rightarrow$ P($\overline A$) = $\frac{5}{6}$ and P($\overline B$) = $\frac{3}{4}$

Also, A and B are independent .A and not B are independent, not A and B are independent.

To Find:

(i) The probability that both of them are selected.

We know that, P( both of them are selected) = P(A ∩ B) = P(A) x P(B)

$\frac{1}{6}\times\frac{1}{4}$

$\frac{1}{24}$

Therefore , The probability that both of them are selected is $\frac{1}{24}$

(ii) P(only one of them is selected) = P(A and not B or B and not A)

= P(A and not B) + (B and not A)

= P( A ∩ $\overline B$) + P(B ∩ $\overline A$) $\frac{1}{3}$

Therefore, the probability that only one of them Is selected is$\frac{1}{3}$

(iii) none is selected

we know that P(none is selected) = P($\overline A\cap \overline B$)

= P($\overline A$) x P($\overline B$)

$\frac{5}{6}\times\frac{3}{4}$

$\frac{5}{8}$

Therefore , the probability that none is selected is$\frac{5}{8}$

(iv) atleast one of them is selected

Now, P(atleast one of them is selected)

= P(selecting only A ) + P(selecting only B) + P(selecting both)

= P(A and not B) + P (B and not A) + P (A and B) $\frac{3}{8}$

Therefore, the probability that atleast one of them is selected is$\frac{3}{8}$