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.
Given : A and B appear for an interview ,then P(A) = 
and P(B) = 
P(
) = 
and P(
) = 
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) 
P(B)
= 
= 
Therefore , The probability that both of them are selected is 
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 
) + P(B 
)
= P(A) 
P(
) + P(B) 
P(
)
= 
+ 
)
= 
+ 
= 
Therefore, the probability that only one of them Is selected is 
iii)none is selected
we know that P(none is selected) = P(
)
= P(
) 
P(
)
= 
= 
Therefore , the probability that none is selected is 
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)
= P( A 
) + P(B 
) + P( A 
B)
= P(A) 
P(
) + P(B) 
P(
) + P(A) 
P(B)
= 
+ 
)+ 
= 
+ 
+ 
= 
Therefore, the probability that atleast one of them is selected is 