Question

$A$ is a set containing $n$ distinct elements. A non-zero subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P$. A non-zero subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ & $Q$ have no common elements

• A. $\frac{(3^n-2^{n+1}+1)}{(2^n-1{)}^2}$
• B. $\frac{(3^n+2^{n+1}+1)}{(4^n-1)}$
• C. $\frac{(3^n+2^{n+1}+1)}{(2^n-1{)}^2}$
• D. $\frac{(3^n{2}^{n+1}+1)}{(2^n-1{)}^2}$

Answer 1

The correct option is A $\frac{(3^n-2^{n+1}+1)}{(2^n-1{)}^2}$

$=\frac{{}^n{C}_1(2^{n-1}-1){+}^n{C}_2(2^{n-2}-1)+.....{+}^n{C}_{n-1}(2-1)}{(2^n-1{)}^2}$
$=\frac{{}^n{C}_1{2}^{n-1}{+}^n{C}_2{2}^{n-2}+.....{+}^n{C}_{n-1}(2^n-2)}{(2^n-1{)}^2}$
$=\frac{(1+2{)}^n-2^n-1-2^n+2}{(2^n-1{)}^2}=\frac{3^n-2^{n+1}+1}{(2^n-1{)}^2}$