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Independent Events

Let X be a se...

Question

$L e t X b e a s e t c o n t a i n i n g n e l e m e n t s . T w o s u b s e t s A a n d B o f X a r e c h o s e n a t r a n d o m, t h e p r o b a b i l i t y t h a t$ $A \cup B = X$ is

\frac{^{2 n} C_{n}}{2^{2 n}}

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\frac{1}{^{2 n} C_{n}}

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1.3.5. . . . . . . \frac{(2 n - 1)}{2^{n} n!}

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(\frac{3}{4})^{n}

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Solution

The correct option is D $(\frac{3}{4})^{n}$
Let $X = x_{1}, x_{2}, . . . . x_{n}$
For each $x_{i} \in X (1 \leq i \leq n)$ , we have the following four choices
$(1) x_{i} \in A a n d x_{i} \in B (2) x_{i} \in A a n d x_{i} / \in B (3) x_{i} / \in A a n d x_{i} / \in B (4) x_{i} / \in A a n d x_{i} / \in B$
Thus the number of ways of choosing A and B is $4^{n}$ . Out of the occurrence of $A \cup B = X$ . Therefore, the number of favourable ways is $3^{n}$ .
Hence, the probability of the required event is ${(\frac{3}{4})}^{n}$

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