Chapter 4.2, Problem 25E

To determine

To find:

The probability that either both dice are sixes or neither die is a six.

Answer to Problem 25E

Solution:

The probability that either both dice are sixes or neither die is a six is $\frac{26}{36}$.

Explanation of Solution

Given:

Two six-sided dice are tossed.

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:

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 that both dice are six.

That is,

$\begin{array}{rll}E& =& (6,6)\\ n(E)& =& 1\end{array}$

$\begin{array}{rll}P(E)& =& \frac{n(E)}{n(S)}\\ P\left(E\right)& =& \frac{1}{36}\end{array}$

Also, $F$ is the event that neither of the dice is six, then,

$P\left(F\right)=\left(1-\frac{11}{36}\right)$

Since, it is impossible to select that both dice are sixes and neither die is a six, so, the probability will be,

$P\left(E\text{and}F\right)=0$

Now, the probability either both dice are sixes or neither die is a six is,

$\begin{array}{rll}P\left(E\text{or}F\right)& =& P\left(E\right)+P\left(F\right)-P\left(E\text{and}F\right)\\ & =& \frac{1}{36}+\left(1-\frac{11}{36}\right)-0\\ & =& \frac{1}{36}+\frac{25}{36}\\ & =& \frac{26}{36}\end{array}$

Therefore, the probability that either both dice are sixes or neither die is a six is $\frac{26}{36}$.