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 P and Q have equal number of elements, is |
$\left(\frac{3}{4}\right)^n$ $\frac{^{2n}C_n}{4^n}$ $\frac{^{2n}C_{n-1}}{4^n}$ $\frac{n^2}{4^n}$ |
$\frac{^{2n}C_n}{4^n}$ |
The set A has n elements. So, it has $2^n$ subsets. Therefore, set P can be chosen in ${^{2n}C}_1$ ways. Similarly, set Q can also be chosen in ${^{2n}C}_1$ ways. Let the subset P of A contains r elements, with ) ≤ r < n. Then, the number of ways of choosing P is ${^nC}_r$. Similarly, Q can be chosen in ${^nC}_r$ ways. So, P and Q can be chosen in ${^nC}_r × {^nC}_r $ ways. But, r can vary from 0 to n. ∴ P and Q can be chosen in $\sum\limits^{n}_{r=0} {^nC}_r × {^nC}_r $ ways, = ${^{2n}C}_n$ ways. ∴ Required probability $=\frac{^{2n}C_n}{4^n}$ |