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Four fair dic...

Question

Four fair dice $D_{1}, D_{2}, D_{3}$ and $D_{4}$ each having six faces numberd $1, 2, 3, 4, 5$ 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

\frac{91}{216}

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\frac{108}{216}

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\frac{125}{216}

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\frac{127}{216}

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Solution

The correct option is A $\frac{91}{216}$
Favourable : $D_{4}$ shows a number and only $1$ of $D_{1}, D_{2}, D_{2}$ shows same number.
or $2$ of $D_{1}, D_{2}, D_{3}$ shows same number
or $3$ of $D_{1}, D_{2}, D_{3}$ shows same number.

Required probability = $\frac{^{6} C_{1} (^{3} C_{1} \times 5 \times 5 +^{3} C_{2} \times 5 +^{3} C_{3})}{216 \times 6}$
= $\frac{6 \times (75 + 15 + 1)}{216 \times 6}$
$= \frac{91}{216}$

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Four fair dice D1,D2,D3 and D4 each having six faces numberd 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

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