Chapter 20, Problem 16E

To determine

To find: The probability model for the total number of spots on two dice.

Answer to Problem 16E

Solution: The probability model for the total number of spots on two dice is given below.

$\begin{array}{cccccccccccc}\text{Total spots}& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ \text{Probability}& \frac{1}{36}& \frac{2}{36}& \frac{3}{36}& \frac{4}{36}& \frac{5}{36}& \frac{6}{36}& \frac{5}{36}& \frac{4}{36}& \frac{3}{36}& \frac{2}{36}& \frac{1}{36}\end{array}$

Explanation of Solution

Calculation:

There are six spots on the six faces of a dice numbered from 1 to 6. So, when two dice are rolled, the possible outcomes or the sample space is as follows:

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

Each of the above outcomes is equally likely with probability $\frac{1}{36}$.

The total number of spots will have the following sample space:

$S_{\text{t}}=\left\{2,3,4,5,6,7,8,9,10,11,12\right\}$

The possible outcomes for each value of the total spots and the associated probabilities are as follows:

$\begin{array}{ccc}\text{Total value}& \text{Possible outcomes}& \text{Probability}\\ 2& \left\{\left(1,1\right)\right\}& \frac{1}{36}\\ 3& \left\{\left(1,2\right),\left(2,1\right)\right\}& \frac{1}{36}+\frac{1}{36}=\frac{2}{36}\\ 4& \left\{\left(1,3\right),\left(3,1\right),\left(2,2\right)\right\}& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{3}{36}\\ 5& \left\{\begin{array}{l}\left(1,4\right),\left(4,1\right),\\ \left(2,3\right),\left(3,2\right)\end{array}\right\}& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{4}{36}\\ 6& \left\{\begin{array}{l}\left(1,5\right),\left(5,1\right),\left(2,4\right),\\ \left(4,2\right),\left(3,3\right)\end{array}\right\}& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{5}{36}\\ 7& \left\{\begin{array}{l}\left(1,6\right),\left(6,1\right),\left(2,5\right),\\ \left(5,2\right),\left(3,4\right),\left(4,3\right)\end{array}\right\}& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{6}{36}\\ 8& \left\{\begin{array}{l}\left(2,6\right),\left(6,2\right),\left(3,5\right),\\ \left(5,3\right),\left(4,4\right)\end{array}\right\}& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{5}{36}\\ 9& \left\{\begin{array}{l}\left(3,6\right),\left(6,3\right),\\ \left(4,5\right),\left(5,4\right)\end{array}\right\}& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{4}{36}\\ 10& \left\{\left(4,6\right),\left(6,4\right),\left(5,5\right)\right\}& \frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{3}{36}\\ 11& \left\{\left(5,6\right),\left(6,5\right)\right\}& \frac{1}{36}+\frac{1}{36}=\frac{2}{36}\\ 12& \left\{\left(6,6\right)\right\}& \frac{1}{36}\end{array}$

Since each outcome is independent, the individual probabilities are summed up.

The above probability model explains the total number of spots that can appear on the up-faces of two dice and the probability associated with them.

To find: The expected value of the total number of spots appearing on two dice.

Solution: The expected value of the total number of spots is 7.

Explanation:

Calculation:

The expected value of any discrete random variable X is computed by using the following formula:

$E\left(X\right)={\displaystyle \sum _{X}x\times p\left(X=x\right)}$

where $p\left(X=x\right)$ denotes the probability that the random variable takes value $x$.

So, by using the probability model obtained above and the formula of expectation, the expected value of the total number of spots is computed as follows:

$\begin{array}{c}E\left(\mathrm{Total}\right)=\left[\begin{array}{l}\left(2\times \frac{1}{36}\right)+\left(3\times \frac{2}{36}\right)+\left(4\times \frac{3}{36}\right)+\left(5\times \frac{4}{36}\right)+\left(6\times \frac{5}{36}\right)+\left(7\times \frac{6}{36}\right)\\ +\left(8\times \frac{5}{36}\right)+\left(9\times \frac{4}{36}\right)+\left(10\times \frac{3}{36}\right)+\left(11\times \frac{2}{36}\right)+\left(12\times \frac{1}{36}\right)\end{array}\right]\\ =\left[\frac{1}{36}\times \left(2+6+12+20+30+42+40+36+30+22+12\right)\right]\\ =\frac{1}{36}\times 252\\ =7\end{array}$