LVI OM - I - Problem 10

Among all subsets of a fixed $ n $-element set $ X $, we sequentially draw with replacement three sets $ A $, $ B $, $ C $. Each time, the probability of drawing any of the $ 2^n $ subsets of set $ X $ is equally likely. Determine the most probable number of elements in the set $ A\cap B\cap C $.
The set of elementary events $\Omega$ consists of all triples $(A,B,C)$, where $A$, $B$, and $C$ are subsets of a given $n$-element set $S$. The selection of each of the $(2^n)^3 = 8^n$ triples $(A,B,C)$ is equally probable. Let $X_k$ $(k = 0,1,2,\ldots,n)$ denote the event that the triple of sets $(A,B,C)$ satisfies the condition $|A \cap B \cap C| = k$, where $|F|$ denotes the number of elements in the set $F$. We seek the natural number $k$ for which the probability of the event $X_k$ is the greatest.

Let us fix a number $k \in \{0,1,2, \ldots,n\}$. We will calculate the number of triples $(A, B, C)$ for which $|A \cap B \cap C| = k$.
If $(A,B,C)$ is a triple of subsets of the set $S$, then each element of the set $S$ is in exactly one of the following eight sets

In order to obtain a triple of sets $(A, B, C)$ for which $|A \cap B \cap C| = k$, we first choose $k$ elements of the set $S$ to place in the set $A \cap B \cap C$ — this can be done in $n \choose k$ ways. Each of the remaining $n-k$ elements of $k$ is assigned one of the remaining seven possibilities listed in (1) — this can be done in $7^{n-k}$ ways. The described procedure uniquely determines the triple $(A,B,C)$. Therefore, the number of triples $(A,B,C)$ for which $|A \cap B \cap C| = k$ is ${n \choose k} \cdot 7^{n-k}$. Hence, we obtain

To determine the number $k$ for which the probability of the event $X_k$ is the greatest, we calculate the ratio

From this, it follows that

Denoting $\lambda = [(n-7)/8]$, where $[x]$ is the greatest integer not greater than $x$, we obtain

Thus, the probability $P(X_k)$ is the greatest for $k = \lambda + 1 = [\frac{1}{8} (n+1)]$.
If $\frac{1}{8}(n-7)$ is an integer, then there are two values of $k$ for which the number $P(X_k)$ is the greatest, namely $k = \frac{1}{8} (n-7)$ and $k = \frac{1}{8} (n+1)$.