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…
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 $\mathrm{D}_{3}$ is
$\frac{91}{216}$
$\frac{108}{216}$
$\frac{125}{216}$
$\frac{127}{216}$
Solution
$D_{4}$ can show a number appearing on one of $D_{1}, D_{2}$ and $D_{3}$ in the following cases.
Case $\mathbf{I}: D_{4}$ shows a number which is shown by exactly one of $D_{1}, D_{2}$ and $D_{3}$.
$D_{4}$ shows a number in ${ }^{6} C_{1}$ ways.
One out of $D_{1}, D_{2}$ and $D_{3}$ can be selected in ${ }^{3} C_{1}$ ways.
[The selected die shows the same number as on $D_{4}$ in one way and rest two dice show the different number in 5 ways each, ]
$\therefore$ Number of ways
$={ }^{6} C_{1} \times{ }^{3} C_{1} \times 1 \times 5 \times 5=450$
Case II : $D_{4}$ shows a number which is shown by exactly two of $D_{1}, D_{2}$ and $D_{3}$.
Number of ways
$={ }^{6} C_{1} \times{ }^{3} C_{2} \times 1 \times 1 \times 5=90$
Case III : $D_{4}$ shows a number which is shown by all three dice $D_{1}, D_{2}$ and $D_{3}$.
Number of ways
$={ }^{6} C_{1} \times{ }^{3} C_{3} \times 1 \times 1 \times 1=6$
$\therefore$ Total number of favourable ways $=450+90+6=546$ Total ways $=6 \times 6 \times 6 \times 6$
$\therefore$ Required Probability $=\frac{546}{6 \times 6 \times 6 \times 6}=\frac{91}{216}$