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:

We are given a set X with 10 elements. The power set P(X) contains all subsets of X, and since each element can be included or excluded independently, the total number of subsets is $$2^{10} = 1024$$.

We pick two sets A and B from P(X) with replacement. This means each pick is independent, and the total number of possible pairs (A, B) is $$|P(X)| \times |P(X)| = 2^{10} \times 2^{10} = 2^{20}$$.

We need the probability that A and B have the same number of elements. Let this common size be k, where k can range from 0 to 10. For each fixed k, the number of subsets with exactly k elements is given by the binomial coefficient $$\binom{10}{k}$$, which is the number of ways to choose k elements from 10.

For a specific k, the number of choices for A with size k is $$\binom{10}{k}$$, and similarly for B. Since the choices are independent, the number of pairs (A, B) where both have size k is $$\binom{10}{k} \times \binom{10}{k} = \left(\binom{10}{k}\right)^2$$.

Since k can be any integer from 0 to 10, we sum over all possible k to get the total number of favorable pairs:

$$\sum_{k=0}^{10} \left(\binom{10}{k}\right)^2$$

There is a known combinatorial identity: $$\sum_{k=0}^{n} \left(\binom{n}{k}\right)^2 = \binom{2n}{n}$$. For n = 10, this becomes:

$$\sum_{k=0}^{10} \left(\binom{10}{k}\right)^2 = \binom{20}{10}$$

Therefore, the total number of favorable pairs is $$\binom{20}{10}$$.

The probability is the number of favorable pairs divided by the total number of possible pairs:

$$\frac{\binom{20}{10}}{2^{20}}$$

Comparing with the options:

• Option A: $$\frac{(2^{10}-1)}{2^{20}}$$
• Option B: $$\frac{\binom{20}{10}}{2^{20}}$$
• Option C: $$\frac{\binom{20}{10}}{2^{10}}$$
• Option D: $$\frac{(2^{10}-1)}{2^{10}}$$

Our result matches option B.

Hence, the correct answer is Option B.