Chapter 4.2, Problem 29E

To determine

To find:

The probability of obtaining either an odd total or a total less than six.

Answer to Problem 29E

Solution:

The probability of obtaining either an odd total or a total less than six is $\frac{22}{36}$.

Explanation of Solution

Given:

Two six-sided dice is rolled.

Formula used:

The addition rule of probability states that, for two events $E$ and $F$, The probability that $E$ or $F$ occurs is given by the formula as follows,

$P\left(E\text{or}F\right)=P\left(E\right)+P\left(F\right)-P\left(E\text{and}F\right)$

Calculation:

Two six-sided dice are rolled.

If two dice are tossed, then, the outcome is,

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

$n(S)=36$

Consider that $E$ is the event of getting an outcome of odd sum when two six-sided dice are rolled, then,

$\begin{array}{rll}P\left(E\right)& =& \frac{n(E)}{n(S)}\\ P\left(E\right)& =& \frac{\left\{\begin{array}{l}\left(1,2\right)\left(1,4\right),\left(1,6\right),\left(2,1\right),\left(2,3\right),\left(2,4\right),\left(3,2\right),\left(3,4\right),\left(3,6\right),\\ \left(4,1\right),\left(4,3\right),\left(4,5\right),\left(5,2\right),\left(5,4\right),\left(5,6\right),\left(6,1\right),\left(6,3\right),\left(6,5\right)\end{array}\right\}}{36}\\ & =& \frac{18}{36}\end{array}$

Also, $F$ is the event of getting sum less than 6, then,

$\begin{array}{rll}P\left(F\right)& =& \frac{n(F)}{n(S)}\\ P\left(F\right)& =& \frac{\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(3,1\right),\left(3,2\right),\left(4,1\right)\right\}}{36}\\ & =& \frac{10}{36}\end{array}$

The probability of odd sum less than 6 will be,

$\begin{array}{rll}P\left(E\text{and}F\right)& =& \frac{\left\{\left(1,2\right),\left(1,4\right),\left(2,1\right),\left(2,3\right),\left(3,2\right),\left(4,1\right)\right\}}{36}\\ & =& \frac{6}{36}\end{array}$

Now, the probability of obtaining either an odd total or a total less than six is,

$\begin{array}{rll}P\left(E\text{or}F\right)& =& P\left(E\right)+P\left(F\right)-P\left(E\text{and}F\right)\\ & =& \frac{18}{36}+\frac{10}{36}-\frac{6}{36}\\ & =& \frac{22}{36}\end{array}$

Therefore, the probability of obtaining either an odd total or a total less than six is $\frac{22}{36}$.