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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
20C10210
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B
(2101)220
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C
20C10220
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D
(2101)210
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Solution

The correct option is C $$\displaystyle \frac{^{20}{C}_{10}}{2^{20}}$$
$$X$$ is a set containing $$10$$ elements.
Then power set $$P(X)$$ contains $$2^{10}$$ elements.
$$A$$ and $$B$$ are picked at random from $$P(X)$$ with replacement.
Total no. of outcomes are $$2^{10}.$$
no. of ways in which $$A$$ and $$B$$ have equal no of elements is $$\displaystyle^{10}C_0^{10}C_0+^{10}C_1^{10}C_1+^{10}C_2^{10}C_2+....+^{10}C_{10}^{10}C_{10}$$
$$\therefore$$ Probability $$=\displaystyle\frac{\displaystyle ^{10}C_0^{10}C_0+^{10}C_1^{10}C_1+^{10}C_2^{10}C_2+....+^{10}C_{10}^{10}C_{10}}{2^{20}}$$
                           $$=\displaystyle\frac{^{20}C_{10}}{2^{20}}$$

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