Question

the probability that two randomly selected subsets of the set have exactly two elements in their intersection

• A. 27/128
• B. 3/16
• C. 1/16
• D. 1/4

Answer 1

Answer: (A)

27/128

Answer 2

Let $S$ be a set with $n$ elements. The probability that a specific element is in a randomly selected subset is $1/2$. For two randomly selected subsets $A$ and $B$, the probability that a specific element is in both $A$ and $B$ (i.e., in $A \cap B$) is $(1/2) \times (1/2) = 1/4$. The number of elements in the intersection, $|A \cap B|$, follows a binomial distribution $Bin(n, 1/4)$. The probability of having exactly $k$ elements in the intersection is $P(|A \cap B|=k) = \binom{n}{k} (1/4)^k (3/4)^{n-k}$.

Since the size of "the set" ($n$) is not specified, we assume $n=4$ as a common convention for such problems. We want to find the probability that $|A \cap B|=2$. Using the binomial probability formula with $n=4$ and $k=2$: $P(|A \cap B|=2) = \binom{4}{2} (1/4)^2 (3/4)^{4-2}$ $P(|A \cap B|=2) = \binom{4}{2} (1/4)^2 (3/4)^2$ $P(|A \cap B|=2) = 6 \times (1/16) \times (9/16)$ $P(|A \cap B|=2) = 6 \times 9 / (16 \times 16)$ $P(|A \cap B|=2) = 54 / 256$ Simplifying the fraction: $P(|A \cap B|=2) = 27 / 128$