Let S be a set containing n elements. If we s | vedclass

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(D) For each element $x \in S$,there are $4$ possibilities for its membership in $A$ and $B$: $1$. $x \in A$ and $x 
otin B$ $2$. $x 
otin A$ and $x

Vedclass · Accepted Answer

$1/2^n$

Vedclass · Answer

(D) For each element $x \in S$,there are $4$ possibilities for its membership in $A$ and $B$: $1$. $x \in A$ and $x 
otin B$ $2$. $x 
otin A$ and $x \in B$ $3$. $x \in A$ and $x \in B$ $4$. $x 
otin A$ and $x 
otin B$ Since there are $n$ elements,the total number of ways to choose two subsets $A$ and $B$ is $4^n = (2^2)^n = (2^n)^2$. For the condition $A \cup B = S$ and $A \cap B = \phi$ to hold,for every element $x \in S$,it must be that either ($x \in A$ and $x 
otin B$) or ($x 
otin A$ and $x \in B$). This means for each of the $n$ elements,there are exactly $2$ choices. Thus,the number of favorable cases is $2^n$. The required probability is $\frac{2^n}{(2^n)^2} = \frac{2^n}{2^{2n}} = \frac{1}{2^n}$.

Let $S$ be a set containing $n$ elements. If we select $2$ subsets $A$ and $B$ of $S$ at random,then the probability that $A \cup B = S$ and $A \cap B = \phi$ is:

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