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  • Let S be the universal set and . The probability of selecting two subsets A 

Let S be the universal set and \mathrm{n(X)=k}. The probability of selecting two subsets A and B of the set X such that \mathrm{B=\bar{A}} is

Option: 1

\frac{1}{2}


Option: 2

\mathrm{\frac{1}{2^k-1}}


Option: 3

\frac{1}{2^{\text{k}}}


Option: 4

\frac{1}{3^{\text{k}}}


Answers (1)

best_answer

The total number of subsets of \mathrm{X \: is\; 2^k. So, n(S)={ }^2{ }^k C_2.}

\mathrm{n(E)=} the number of selections of two nonintersecting subsets whose union is X.

           \mathrm{=\frac{1}{2}\left({ }^k C_0+{ }^k C_1+{ }^k C_2+\ldots\right)}

           ( \because  the number of selections in which one subset has r elements and the rest are in the other subset \mathrm{={ }^k C_r} and every selection             appears twice in the total number of selections).

\mathrm{\therefore \quad P(E)=\frac{(1 / 2) \cdot 2^k}{2^k C_2}=\frac{2^{k-1}}{\frac{2^k\left(2^k-1\right)}{2}}=\frac{1}{2^k-1} .}

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