91Ó°ÊÓ

A fair coin is tossed three times.

The fair coins are tossed. If one die toss the possible outcomes are \(\left\{ {1,2,3,4,5,6} \right\}\)

Here 3 fair dice are rolled

So the total possible outcomes are \({6^3} = 216\)

\(n\left( S \right) = 216\)

Here, we need to find the probability of all three faces will be same.

Let A be a event that represent the all three faces will be same

The favourable outcomes for event A is

\(A = \left\{ {\left( {1,1,1} \right),\left( {2,2,2} \right),\left( {3,3,3} \right),\left( {4,4,4} \right),\left( {5,5,5} \right),\left( {6,6,6} \right)} \right\}\)

\(n\left( A \right) = 6\)

The probability of all three faces is same is

\(\begin{aligned}{c}P\left( A \right) &= \frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{Favourable}}\,{\bf{outcomes}}}}{{{\bf{Total}}\,{\bf{outcomes}}}}\\ &= \frac{{{\bf{n}}\left( {\bf{A}} \right)}}{{{\bf{n}}\left( {\bf{S}} \right)}}\\ &= \frac{6}{{216}}\\ &= 0.027778\end{aligned}\)

Thus, the probability that all three faces will be is 0.027778.