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(C) Let $S = \{1, 2, 3, 4, 5\}$. The total number of subsets of $S$ is $2^{5} = 32$.Since two subsets $A$ and $B$ are selected randomly,the total numb

Vedclass · Accepted Answer

$\frac{135}{2^{9}}$

Vedclass · Answer

(C) Let $S = \{1, 2, 3, 4, 5\}$. The total number of subsets of $S$ is $2^{5} = 32$.Since two subsets $A$ and $B$ are selected randomly,the total number of pairs $(A, B)$ is $32 	imes 32 = 2^{10}$.For each element $x \in S$,there are $4$ possibilities for its membership in $A$ and $B$: $x 
otin A, x 
otin B$; $x \in A, x 
otin B$; $x 
otin A, x \in B$; or $x \in A, x \in B$.We want the intersection $A \cap B$ to have exactly $2$ elements.First,choose $2$ elements out of $5$ to be in the intersection: $\binom{5}{2} = 10$ ways.For the remaining $3$ elements,each must not be in the intersection,meaning they can be in $(A \setminus B)$,$(B \setminus A)$,or in neither $A$ nor $B$. There are $3$ such choices for each of the $3$ remaining elements,giving $3^{3} = 27$ ways.Thus,the number of favorable pairs is $10 	imes 27 = 270$.The probability is $\frac{270}{2^{10}} = \frac{270}{1024} = \frac{135}{512} = \frac{135}{2^{9}}$.

The probability that two randomly selected subsets of the set $\{1, 2, 3, 4, 5\}$ have exactly two elements in their intersection is:

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