BITSAT 2005 Solved Question Paper

Let $A$ and $B$ be two subsets of $S$ . There are following cases to make a subset of $S$, under the given condition i.e. $A \cup B = S$ and $A \cap B = \phi$ Case I : If set $A$ has no element. The number of ways of selection of 0 element from set $S$ is $\binom{n}{0}$ . Case II : If set A has one element. The number of ways of selection of one element from set $S$ is $\binom{n}{1}$ . Case III : If set A has two elements. The number of ways of selection of two element from set $S$ is $\binom{n}{2}$ . .................................................................... .................................................................... ....................................................................Case $(n)$ : If set $A$ has $n$ elements. The number of ways of selection of $n$ elements from set $S$ is " $\binom{n}{n}$ . $\therefore$ Total set of $A = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n}$ $=2^{n}$ Total set of $A$ arid $B = 2^{n} \times 2^{n} = 2^{2n}$ Required probability $=\frac{2^{n}}{2^{2n}}=\frac{1}{2^{n}}$