The student mess committee of a reputed Engineering College has n members. Let P be the event that the Committee has students of both sexes and let Q be the event that there is at most one female student in the committee. Assuming that each committee member has probability 0.5 of being female, the value of n for which the events P and Q are independent is

Let's first estimate the sample space. In a committee of n members, there can be the following cases:

(0 males, n females), (1 male, n-1 females), (2 males, n-2 females) ...(n males, 0 females).

Thus, the sample space has n+1 cases.

n(S) = n+1 cases

Barring 2 cases from the sample space, event P occurs for all other cases. 

=> n(P) = n+1-2 = n-1 cases

=> p(P) = (n-1)/(n+1)

At most 1 female can be expressed as the sum of the cases exactly 0 females and exactly 1 female.

=> n(Q) = 2 cases

=> p(Q) = 2/(n+1)

The set $$P \cap Q$$ occurs when there is exactly one female member.

Hence, n($$P \cap Q$$) = 1

p($$P \cap Q$$) = 1/(n+1)

When two events are independent, the probability of both of them occurring is equal to the product of their individual probabilities.

Thus,

p($$P \cap Q$$) = p(P) * p(Q)

1/(n+1) = (n-1)/(n+1) * 2/(n+1)

=> n+1 = 2n -2

=> n= 3