**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\(\frac{3}{8}\)