Four fair dice $D_1$, $D_2$, $D_3$ and $D_4$; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that $D_4$ shows a number appearing on one of $D_1$, $D_2$ and $D_3$ is

Four fair dice D₁, D₂, D₃ and D₄; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that D₄ shows a number appearing on one of D₁, D₂ and D₃ is

$\text{Four dice } D_1, D_2, D_3, D_4 \text{ rolled simultaneously; each face } 1\text{–}6$

$\text{Total outcomes} = 6^4 = 1296$

$\text{Event } A:\ D_4 \text{ shows a number appearing on at least one of } D_1, D_2, D_3$

$\text{To find: } P(A)$

$\text{Plan: condition on } (D_1, D_2, D_3);\ \text{let } S = \text{set of their values}$

$D_4 \text{ matches in } |S| \text{ ways}$

$\text{Count triples by } |S| = 1, 2, 3$

$S = \{\text{distinct values among } d_1, d_2, d_3\}$

$\#\{\text{matching } d_4\} = |S| \Rightarrow P(A) = \dfrac{\sum_{(d_1,d_2,d_3)} |S|}{6^4}$

$|S| = 3:\ \text{all three distinct}$

$|S| = 2:\ \text{exactly two distinct}$

$|S| = 1:\ \text{all three equal}$

$|S|=3:\quad 6\times 5\times 4 = 120$

$\text{contribution } 3 \times 120 = 360$

$|S|=2:\quad \binom{6}{2}\,(2^3 - 2) = 15 \times 6 = 90$

$\text{contribution } 2 \times 90 = 180$

$|S|=1:\quad 6 \times 1 = 6$

$\text{contribution } 1 \times 6 = 6$

$120 + 90 + 6 = 216 = 6^3\ \checkmark$

$\sum |S| = 360 + 180 + 6 = 546$

|S| #triples matches favourable hline 3 120 3 360 2 90 2 180 1 6 1 6 hline 216 546

$P(A) = \dfrac{\sum |S|}{6^4} = \dfrac{546}{1296}$

$= \dfrac{546/6}{1296/6} = \dfrac{91}{216}$

$P(A) = \dfrac{91}{216}\ \Rightarrow\ \text{option (1)}$