MCQ
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
  • $\frac{91}{216}$
  • B
    $\frac{108}{216}$
  • C
    $\frac{125}{216}$
  • D
    $\frac{127}{216}$

Answer

Correct option: A.
$\frac{91}{216}$
a
Favourable: $D _4$ 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

$\text { Required Probability }  =\frac{{ }^6 C _1\left({ }^3 C _1 \times 5 \times 5+{ }^3 C _2 \times 5+{ }^3 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}$

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