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

Solution :
$${D_4}$$ can show a number appearing on one of $${D_1},$$ $${D_2}$$ and $${D_3}$$ in the following cases.
Case I : $${D_4}$$ shows a number which is shown by only 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.
∴ Number of ways to happen case I
$$ = {\,^6}{C_1} \times {\,^3}{C_1} \times 1 \times 5 \times 5 = 450$$
Case II : $${D_4}$$ shows a number which is shown by only two of $${D_1},$$ $${D_2}$$ and $${D_3}.$$
As discussed in case I, it can happen in the following 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 it can be done
$$ = {\,^6}{C_1} \times {\,^3}{C_3} \times 1 \times 1 \times 1 = 6$$
∴ Total number of favourable ways = 450 + 90 + 6 = 546
Also total ways $$ = 6 \times 6 \times 6 \times 6$$
∴ Required Probability $$ = \frac{{546}}{{6 \times 6 \times 6 \times 6}} = \frac{{91}}{{216}}$$