91Ó°ÊÓ

Here two balanced dice are rolled.

Let S represent the sample space of the experiment.

\(S = \left\{ \begin{array}{l}\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right)\\\,\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right)\\\,\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {3,6} \right)\\\,\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right),\left( {4,6} \right)\\\,\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right),\left( {5,6} \right)\\\,\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {6,6} \right)\end{array} \right\}\)

In the sample space, the possible outcomes are 36

In the sample space, there are a total of 36 pairs. of observation in which 18 pairs produce sum of two numbers is odd.

So the number offavourable outcomes is 18

\(\begin{array}{c}\Pr \left( {Sum\,of\,two\,numbers\,appear\,odd} \right) = \frac{{{\bf{Number}}\,{\bf{of}}\,{\bf{Favourable}}\,{\bf{outcomes}}}}{{{\bf{Total}}\,{\bf{outcomes}}}}\\ = \frac{{18}}{{36}}\\ = 0.50\end{array}\)

So the probability of the sum of two numbers appearing odd on dice is 0.50