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14.Probability
normal
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
A
$\frac{91}{216}$
B
$\frac{108}{216}$
C
$\frac{125}{216}$
D
$\frac{127}{216}$
(IIT-2012)
Solution
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}$
Standard 11
Mathematics
