Four fair dice $\mathrm{D}_1, \mathrm{D}_2, \mathrm{D}_3$ and $\mathrm{D}_4$ each having six faces numbered $1,2,3,4,5$ and 6 are rolled simultaneously. The probability that $\mathrm{D}_4$ shows a number appearing on one of $\mathrm{D}_1, \mathrm{D}_2$ and $\mathrm{D}_3$ is
Select the correct option:
A
$\frac{91}{216}$
B
$\frac{108}{216}$
C
$\frac{125}{216}$
D
$\frac{127}{216}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Favourable: $\quad \mathrm{D}_4$ shows a number and only 1 of $D_1 D_2 D_3$ shows same number or only 2 of $D_1 D_2 D_3$ shows same number or all 3 of $D_1 D_2 D_3$ shows same number
$$
\begin{aligned}
\text { Required Probability } & =\frac{{ }^6 \mathrm{C}_1\left({ }^3 \mathrm{C}_1 \times 5 \times 5+{ }^3 \mathrm{C}_2 \times 5+{ }^3 \mathrm{C}_3\right)}{216 \times 6} \\
& =\frac{6 \times(75+15+1)}{216 \times 6} \\
& =\frac{6 \times 91}{216 \times 6} \\
& =\frac{91}{216}
\end{aligned}
$$
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