Chapter SH, Problem 13.8.5EP

To determine

To find: probability of getting an odd sum.

Answer to Problem 13.8.5EP

The required probability is $\frac{1}{2}$ .

Explanation of Solution

Given:two six-sided dice are rolled.

Calculation:

Possible number of outcomes in rolling two dice are

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

Total number of outcomes is

  $6\times 6=36$

Outcomes that have odd sum are

  $\begin{array}{l}\left(1,2\right),\left(1,4\right),\left(1,6\right),\left(2,1\right),\left(2,3\right),\left(2,5\right),\hfill \\ \left(3,2\right),\left(3,4\right),\left(3,6\right),\left(4,1\right),\left(4,3\right),\left(4,5\right),\hfill \\ \left(5,2\right),\left(5,4\right),\left(5,6\right),\left(6,1\right),\left(6,3\right),\left(6,5\right).\hfill \end{array}$

Total number of favorable outcomes is 18.

Hence, the required probability is $\frac{18}{36}=\frac{1}{2}$ .

Conclusion:

Therefore, the required probability is $\frac{1}{2}$ .