Practicing Success
Let A be a set consisting of n elements. The probability of selecting two subsets P and Q of set A such that Q = $\overline{P}$ is |
$\frac{1}{2}$ $\frac{1}{2^n-1}$ $\frac{1}{2^n}$ $\frac{1}{3^n}$ |
$\frac{1}{2^n-1}$ |
The set A has $2^n$ elements. Therefore, two subsets P and Q can be chosen in ${^{2n}C}_2$ ways. Suppose P consists of r elements. Then, P can be chosen in ${^nC}_r$. Since, $Q = \overline{P}$. Therefore, P and Q can be chosen in ${^nC}_r$ ways. But, r can vary from 0 to n and P and Q can be interchanged also. ∴ Number of ways of selecting P and Q such that $Q = \overline{P}$ is $\frac{1}{2}\sum\limits^{n}_{r=0} {^nC}_r= 2^{n-1}$ Hence, required probability $=\frac{2^n-1}{^{2n}C_2}=\frac{1}{2^n-1}$ |