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

Engineering

Mathematics

Conditional Probability

Question

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

$\frac{127}{216}$

$\frac{91}{216}$

$\frac{108}{216}$

$\frac{125}{216}$

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Case I : D₄ and D₁ (or D₂ or D₃) show same digit, then number of ways = 6 × 5 × 5 × 3

Case II : D₄ and D₁, D₂ (or D₂D₃ or D₃D₁) show same digit , then number of ways = 6 × 5 × 3

Case III : D₁, D₂, D₃, D₄ all show same digit, then number of ways = 6

$\Rightarrow P = \frac{450 + 90 + 6}{6^{4}} = \frac{91}{216}$ Ans.

Aliter: P (D₃ show one of D₁, D₂, D₃ number) = P(All three D₁, D₂, D₃ shows same number)

+ P (Any two of D₁, D₂, D₃ shows same number and third shows different)

+ P (All three D₁, D₂, D₃ shows different number)

$=^{6} C_{1} {(\frac{1}{6})}^{3} (\frac{1}{6}) +^{6} C_{2} \cdot^{3} C_{2} \cdot 2! {(\frac{1}{6})}^{3} (\frac{2}{6}) +^{6} C_{3} \cdot 3! {(\frac{1}{6})}^{3} (\frac{3}{6})$

= $\frac{1}{6^{4}}$ [6 + 15 ×12 + 6 × 60] = $\frac{91}{216}$ . Ans.