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, is
Answer & explanation
Correct answer: option 4
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.
∴ Sets P and Q can be chosen in ${^{2n}C}_1 × {^{2n}C}_1= 2^n ×2^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 ${^nC}_r × {^nC}_{r+1}$, where 0 ≤ r ≤n-1.
Thus, the number of ways of choosing P and Q in general, is
$\sum\limits^{n-1}_{r=0} {^nC}_r × {^nC}_{r+1} = {^{2n}C}_{n-1}$
Hence, required probability $= \frac{^{2n}C_{n-1}}{4^n}$