Mathematics - Probability Question with Solution | TestHub

$D_4$ 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 $=\frac{546}{6\times 6\times 6\times 6}=\frac{91}{216}$