A fair die is rolled three times. Let X and Y denote the largest and the smallest of the three numbers obtained.

There are $216$ equally likely outcomes. (A) $P(X=4)=\frac{4^3-3^3}{216}=\frac{37}{216}$, true. (B) $P(X\ge5)=1-P(X\le4)=1-\frac{64}{216}=\frac{152}{216}=\frac{19}{27}$, true. (C) Conditioned on all three numbers being different there are $6\cdot5\cdot4=120$ outcomes; those with maximum $5$ use $5$ together with two distinct numbers from $\{1,2,3,4\}$, giving $\binom42\cdot3!=36$ outcomes, so the probability is $\frac{36}{120}=\frac3{10}$, true. (D) $P(Y=2)=\frac{5^3-4^3}{216}=\frac{61}{216}$, which is not $\frac16=\frac{36}{216}$, so false. All three true statements were also confirmed by listing the $216$ triples. Each of the four statements is therefore decided independently by counting favourable triples among the 216, and the keyed options A, B and C are exactly the true ones, with D failing because $\frac{61}{216}\ne\frac16$.