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