InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q is again chosen at random. The probability that Q contains just one element more than P, isA. `(.^(2n)C_(n))/(4^(n))`B. `(2n)/(4^(n))`C. `(.^(2n-1)C_(n))/(4^(n))`D. `(.^(2n)C_(n-1))/(4^(n))` |
|
Answer» Correct Answer - D The set A has n elements. So, it has `2^(n)` subsets. Therefore, set P can be chosen in `2^(n)C_(1)` ways. Similarly, set Q can also be chosen in `2^(n) C_(1)` ways. `therefore` Sets P and Q can be chosen in `.^(2n)C_(1)xx .^(2n)C_(1)=2^(n)xx2^(n)=4^(n)` ways. If P contains r elements, then Q must contain (r+1) elements. In this case the number of ways of choosing P and Q is `.^(n)C_(r )xx .^(n)C_(r+1)`, where `0 le r le n -1`. Thus, the number of ways of choosing P and Q in general, is `underset(r=0)overset(n-1)sum .^(n)C_(r )xx .^(n)C_(r+1)= .^(2n)C_(n-1)` Hence, required probability `=(.^(2n)C_(n-1))/(4^(n))` |
|