A number is chosen at random from \1,2,…,60\. Let A, B, C be the events that the number is divisible by 2, by 3 and by 4 respectively.

Counting in $\{1,\dots,60\}$: $n(A)=30$, $n(B)=20$, $n(C)=15$, so $P(A)=\frac12$, $P(B)=\frac13$, $P(C)=\frac14$. The number is in $A\cap B$ exactly when it is a multiple of $6$, so $P(A\cap B)=\frac{10}{60}=\frac16=P(A)P(B)$, and $A,B$ are independent (A true). Every multiple of $4$ is even, so $C\subset A$ and $P(A\cap C)=\frac14$, whereas $P(A)P(C)=\frac18$; hence $A,C$ are not independent (B false). Because $C\subset A$, $A\cup B\cup C=A\cup B$ and $P=\frac12+\frac13-\frac16=\frac23\neq\frac34$ (C false). Pairwise independence would need all three pairs to be independent, but the pair $A,C$ fails, so D is false (note $B,C$ are independent since multiples of $12$ number $5=\frac{20\cdot15}{60}$). Only the statement in option A is correct; a brute-force enumeration over the sixty outcomes confirms each count.