Chapter 4.3, Problem 59E

(a)

To determine

All 36 pairs of up-faces.

Answer to Problem 59E

Solution: All possible pairs are:

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

Explanation of Solution

In this study, two dice are rolled, each pair has an equal chance to come up. There are 36 possible outcomes which are obtained as:

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

(b)

To determine

The probability of each pair.

Answer to Problem 59E

Solution: The probability of each pair is $1/36$.

Explanation of Solution

There are 36 possible outcomes and each pair has an equal chance to come up. In this case, sample space is 36. Thus, the probability is calculated as:

Probability for pair (1, 1) is:

$\frac{\text{Occurence of}\left(1,1\right)}{\text{Sample space}}=\frac{1}{36}$

Probability of (1, 2) is:

$\frac{\text{Occurence of}\left(1,2\right)}{\text{Sample space}}=\frac{1}{36}$

Now, it can be continued for remaining outcomes. Hence, the probability of each pair is $1/36$.

(c)

To determine

To find: The values of X and give the probability distribution of X. and drawing a probability histogram.

Answer to Problem 59E

Solution: In this study, sum of up faces is defined as X. The probability distribution is:

$\begin{array}{cccccccccccc}\text{sum}& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ \text{probability}& 1/36& 2/36& 3/36& 4/36& 5/36& 6/36& 5/36& 4/36& 3/36& 2/36& 1/36\end{array}$

Explanation of Solution

In the study, there are 36 possible outcomes. The random variable X is represented as sum of up faces. There is one pair whose sum is equal to 2. Then, the probability is $1/36$. There are two pairs whose sum is equal to 3. Then, the probability is $2/36$. Continue this procedure, until the sum of up faces is obtained as 12. Hence, these probabilities can be defined as probability distribution:

$\begin{array}{cccccccccccc}\text{sum}& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ \text{probability}& 1/36& 2/36& 3/36& 4/36& 5/36& 6/36& 5/36& 4/36& 3/36& 2/36& 1/36\end{array}$

Graph: The histogram for the above probability distribution is provided below:

(d)

To determine

The probability of rolling 7 or an 11.

Answer to Problem 59E

Solution: The probability is 0.221.

Explanation of Solution

In the study, there are 6 pairs whose sum is equal to seven, that is, (1,6), (2,5), (3,4), (4,3),(5,2), and (6,1). Also there are 2 pairs whose sum is equal to 11, that is, $\left(5,6\right),\left(6,5\right)$. The probability $P\left[X=7\right]$ is $6/36$ and the probability $P\left[X=11\right]$ is $2/36$. Now, the probability $P\left[7\text{ or 11}\right]$ is calculated as:

$\begin{array}{c}P\left[7\text{ or 11}\right]=P\left[X=7\right]+P\left[X=11\right]\\ =\frac{6}{36}+\frac{2}{36}\\ =0.166+0.055\\ =0.221\end{array}$

Hence, the probability that 7 or 11 comes up on the next roll of two dice is 0.221.

(e)

To determine

The probability that any other sum except a 7 is rolled.

Answer to Problem 59E

Solution: The probability is 0.834.

Explanation of Solution

In the study, there are 6 pairs whose sum is equal to seven, that is, (1,6), (2,5),(3,4), (4,3),(5,2), and (6,1).The probability $P\left[X=7\right]$ is $6/36$. Now, the probability $P\left[\text{any other sum except }7\right]$ calculated as:

$\begin{array}{c}P\left[\text{any other sum except }7\right]=1-P\left(7\right)\\ =1-0.166\\ =0.834\end{array}$

Hence, the probability is 0.834.