Four fair dice D_ 1 , D_ 2 , D_ 3 and D_ 4 ; each having six faces...

$D_{4}$ 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 $=\frac{546}{6 \times 6 \times 6 \times 6}=\frac{91}{216}$