Chapter 2.9, Problem 119E

Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtain

1. a a sum of 3 before we obtain a sum of 7?
2. b a sum of 4 before we obtain a sum of 7?

To determine

Find the probability that a sum of 3 before people obtain a sum of 7.

Answer to Problem 119E

The probability that a sum of 3 before people obtain a sum of 7 is 0.25.

Explanation of Solution

Calculation:

Let A denotes a sum is 3 and B denotes a sums are not 3 or not 7

The possible outcomes when two dice are tossed.

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

Total number of outcomes is 36.

The list of outcomes for getting a sum of 3 is {(1,2), (2,1)}. Therefore, the number of ways of getting a sum of 3 is 2.

The list of outcomes for getting a sum of 7 is {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. Therefore, the number of ways of getting a sum of 7 is 6.

The probability of getting an event A is $\frac{2}{36}$.

The probability of getting an event B is $\frac{28}{36}\left(=1-\frac{2}{36}-\frac{6}{36}\right)$.

Here, a sum of 3 before sum of 7 can happen on the first roll, second roll, third roll, etc. This can be written as, A, BA, BBA, BBBA, etc

The probability that a sum of 3 before people obtain a sum of 7 is obtained as follows:

$\begin{array}{c}\text{Required probability}=P\left(A\right)+P\left(B\right)P\left(A\right)+P\left(B\right)P\left(B\right)P\left(A\right)+\mathrm{...}\\ =\frac{2}{36}+\frac{28}{36}\frac{2}{36}+\frac{28}{36}\frac{28}{36}\frac{2}{36}+\mathrm{...}\\ =\frac{2}{36}+\frac{28}{36}\frac{2}{36}+{\left(\frac{28}{36}\right)}^2\frac{2}{36}+\mathrm{...}\langle \because \text{It represents geometric series}\rangle \\ =\frac{\frac{2}{36}}{1-\frac{28}{36}}\langle \begin{array}{l}{\text{If a+ar+ar}}^2+\mathrm{...}\\ \text{Then sum S}=\frac{a}{1-r}\end{array}\rangle \\ =\frac{2}{8}\\ =0.25\end{array}$

Thus, the probability that a sum of 3 before people obtain a sum of 7 is 0.25.

To determine

Find the probability that a sum of 4 before people obtain a sum of 7.

Answer to Problem 119E

The probability that a sum of 4 before people obtain a sum of 7 is 0.3333.

Explanation of Solution

Calculation:

Let A denotes a sum is 4 and B denotes a sums are not 4 or not 7

The list of outcomes for getting a sum of 4 is {(1,3),(2,2) (3,1)}. Therefore, the number of ways of getting a sum of 3 is 3.

The list of outcomes for getting a sum of 7 is {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. Therefore, the number of ways of getting a sum of 7 is 6.

The probability of getting an event A is $\frac{3}{36}$.

The probability of getting an event B is $\frac{27}{36}\left(=1-\frac{3}{36}-\frac{6}{36}\right)$.

Here, a sum of 4 before sum of 7 can happen on the first roll, second roll, third roll, etc. This can be written as, A, BA, BBA, BBBA, etc

The probability that a sum of 4 before people obtain a sum of 7 is obtained as follows:

$\begin{array}{c}\text{Required probability}=P\left(A\right)+P\left(B\right)P\left(A\right)+P\left(B\right)P\left(B\right)P\left(A\right)+\mathrm{...}\\ =\frac{3}{36}+\frac{27}{36}\frac{3}{36}+\frac{27}{36}\frac{27}{36}\frac{3}{36}+\mathrm{...}\\ =\frac{3}{36}+\frac{27}{36}\frac{3}{36}+{\left(\frac{27}{36}\right)}^2\frac{3}{36}+\mathrm{...}\langle \because \text{It represents geometric series}\rangle \\ =\frac{\frac{3}{36}}{1-\frac{27}{36}}\langle \begin{array}{l}{\text{If a+ar+ar}}^2+\mathrm{...}\\ \text{Then sum S}=\frac{a}{1-r}\end{array}\rangle \\ =\frac{3}{9}\\ =0.3333\end{array}$

Thus, the probability that a sum of 3 before people obtain a sum of 7 is 0.3333.