Chapter 9.7, Problem 29E

To determine

To find the probability of getting a sum of odd and not more than 7.

Answer to Problem 29E

The probability of getting a sum of odd and not more than 7 is $\frac{1}{3}$ .

Explanation of Solution

Given:

A six sided die is tossed twice.

Calculation:

The sample of the experiment of tossing a six sided die twice is,

  $S=\left\{\begin{array}{}$ \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)$\end{array}\right\}$

Now let J is the event of getting a sum of odd and not more than 7.

Sum of odd and not more than 7 means, the sum must be 3, 5 or 7.

  $\therefore J=\left\{\left(1,2\right),\left(1,4\right),\left(1,6\right),\left(2,1\right),\left(2,3\right),\left(2,5\right),\left(3,2\right),\left(3,4\right),\left(4,1\right),\left(4,3\right),\left(5,2\right),\left(6,1\right)\right\}$

Hence the probability of getting a sum of odd and not more than 7 is,

  $\begin{array}{l}P\left(J\right)=\frac{n\left(J\right)}{n\left(S\right)}\\ \Rightarrow P\left(J\right)=\frac{12}{36}\\ \Rightarrow P\left(J\right)=\frac{1}{3}\end{array}$