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Hint- Write samples of cases of total favorable and that of event to reach up to the probability.

If we toss two different dice than, total favorable cases are

$S = \left[ \begin{gathered}

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

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

................................................... \\

...................................\left( {6,5} \right),\left( {6,6} \right) \\

\end{gathered} \right]$

Thus $n(S) = 36$

Now the favorable outcome to get product of 6 are

$\left[ {\left( {1,6} \right),\left( {6,1} \right),\left( {2,3} \right),\left( {3,2} \right)} \right]$

Thus let E be an event of getting a product 6 on toss of two dice then $n(E) = 4$

Now $P(E) = \dfrac{{Favorable{\text{ cases of event E}}}}{{Total{\text{ possible sample cases}}}} = \dfrac{{n(E)}}{{n(S)}}$……………………………. (1)

Using equation (1)

$P(E) = \dfrac{4}{{36}} = \dfrac{1}{9}$

Hence probability of getting a product 6 while toss of two different dice is $\dfrac{1}{9}$

Note- Whenever we have to solve such type of problems always write down the set of all possible sample cases and then the cases corresponding to that particular event, this helps in reducing the chances of leaving any sample case .Then use the probability basics of $P(E) = \dfrac{{Favorable{\text{ cases of event E}}}}{{Total{\text{ possible sample cases}}}} = \dfrac{{n(E)}}{{n(S)}}$to reach to the solution.

If we toss two different dice than, total favorable cases are

$S = \left[ \begin{gathered}

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

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

................................................... \\

...................................\left( {6,5} \right),\left( {6,6} \right) \\

\end{gathered} \right]$

Thus $n(S) = 36$

Now the favorable outcome to get product of 6 are

$\left[ {\left( {1,6} \right),\left( {6,1} \right),\left( {2,3} \right),\left( {3,2} \right)} \right]$

Thus let E be an event of getting a product 6 on toss of two dice then $n(E) = 4$

Now $P(E) = \dfrac{{Favorable{\text{ cases of event E}}}}{{Total{\text{ possible sample cases}}}} = \dfrac{{n(E)}}{{n(S)}}$……………………………. (1)

Using equation (1)

$P(E) = \dfrac{4}{{36}} = \dfrac{1}{9}$

Hence probability of getting a product 6 while toss of two different dice is $\dfrac{1}{9}$

Note- Whenever we have to solve such type of problems always write down the set of all possible sample cases and then the cases corresponding to that particular event, this helps in reducing the chances of leaving any sample case .Then use the probability basics of $P(E) = \dfrac{{Favorable{\text{ cases of event E}}}}{{Total{\text{ possible sample cases}}}} = \dfrac{{n(E)}}{{n(S)}}$to reach to the solution.

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