Question

Let $X$ be a set containing $10$ elements and $P(X)$ be its power set. If $A$ and $B$ are picked up at random from $P(X)$, with replacement, then the probability that $A$ and $B$ have equal number of elements, is :

• A. $\frac{^{20}C_{10}}{2^{10}}$
• B. $\frac{\left(2^{10}-1\right)}{2^{20}}$
• C. $\frac{\left(2^{10}-1\right)}{2^{10}}$
• D. $\frac{^{20}C_{10}}{2^{20}}$

Answer 1

Answer: (D)

$n(x)=10, n(P(x))=(2)^{10}= 1024$
$A, B \in P(x)$
$n$ (sample space) $=2^{10} \times 2^{10}$
$n(E) =\left(1 \times 1+{ }^{10} C_{1} \times{ }^{10} C_{1}+{ }^{10} C_{2} \times{ }^{10} C_{2}+{ }^{10} C_{3} \times{ }^{10} C_{3}+\ldots+{ }^{10} C_{10} \times{ }^{10} C_{10}\right)$
$=\left({ }^{10} C_{0}\right)^{2}+\left({ }^{10} C_{1}\right)^{2}+\left({ }^{10} C_{2}\right)^{2}+\ldots+\left({ }^{10} C_{10}\right)^{2}$
$={ }^{20} C_{10}$
$\therefore P(E) =\frac{{ }^{20} C_{10}}{2^{10} \times 2^{10}}=\frac{{ }^{20} C_{10}}{2^{20}}$