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

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