Chapter 13, Problem 13.48E

(a)

To determine

To find: The probability of rolling doubles on a single toss of the dice.

Answer to Problem 13.48E

The probability of rolling doubles on a single toss of the dice is 0.167.

Explanation of Solution

Given info:

A rolling pair of balanced dice in a board game is given and the rolls are independent of each other.

Calculation:

Two dice are rolled the sample space S is given as follows:

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

The probability of rolling doubles on a single toss of the dice is obtained below:

The possible outcomes to get doublets are $\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\right\}$

The number of double outcomes in a pair of dice is 6 and the total number of outcomes is 36.

The required probability is as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Rolling}\text{doubles}\\ \text{in a}\text{single}\text{dice}\end{array}\right)=\frac{\text{Number of favorable cases}}{\text{Total number of cases}}\\ =\frac{6}{36}\\ =\frac{1}{6}\\ =0.167\end{array}$

Thus, the probability of rolling doubles on a single toss of the dice is 0.167.

(b)

To determine

To find: The probability of getting no doubles on the first toss, but doubles on the second toss.

Answer to Problem 13.48E

The probability of getting no doubles on the first toss, but doubles on the second toss is0.139.

Explanation of Solution

Calculation:

From part (a), the probability of rolling doubles on a single toss of the dice is 0.167.

Hence, the probability of not rolling doubles on a single toss of the dice is $\text{0}\text{.833}\left(=1-0.167\right)$0.167.

Also, each toss is independent of the other.

The required probability is,

$\begin{array}{c}P\left(\begin{array}{l}\text{No}\text{doubles on the first toss and }\\ \text{roll doubles on the second toss}\end{array}\right)=\left[\begin{array}{l}P\left(\text{No}\text{doubles on the first toss}\right)\\ P\left(\text{Roll doubles on the second toss}\right)\end{array}\right]\\ =0.833\times 0.167\\ =\text{0}\text{.139}\end{array}$

Thus, the probability of getting no doubles on the first toss, but doubles on the second toss is0.139.

(c)

To determine

To find: The probability of getting no doubles in first two tosses and getting doubles in the third toss.

Answer to Problem 13.48E

The probability of getting no doubles in first two tosses and getting doubles in the third toss is 0.1159.

Explanation of Solution

Calculation:

The required probability is,

$\begin{array}{c}P\left(\begin{array}{l}\text{No}\text{doubles on the first and second toss, and }\\ \text{roll doubles on the third toss}\end{array}\right)=\left[\begin{array}{l}P\left(\text{No}\text{doubles on the first toss}\right)\\ P\left(\text{No}\text{doubles on the second toss}\right)\\ P\left(\text{Roll doubles on the third toss}\right)\end{array}\right]\\ =0.833\times 0.833\times 0.167\\ =\text{0}\text{.1159}\end{array}$

Thus, the probability of getting no doubles in first two tosses and getting doubles in the third toss is 0.1159.

(d)

To determine

To find: The probability that first doubles occurs on the fourth toss and on the fifth toss and also give the general result that the first doubles occurs on the kth toss.

Answer to Problem 13.48E

The probability that first doubles occurs on the fourth toss is 0.0965.

The probability that first doubles occurs on the fourth toss is 0.0804.

The probability that first doubles occurs on the k^th toss is ${\left(0.833\right)}^{k-1}\left(0.167\right)$.

Explanation of Solution

Given info:

Calculation:

The probability that first doubles occurs on the fourth toss is obtained below:

$\begin{array}{c}P\left(\text{Doubles on the fourth toss}\right)=\left[\begin{array}{l}P\left(\text{No}\text{doubles on the first toss}\right)\\ P\left(\text{No}\text{doubles on the second toss}\right)\\ P\left(\text{No}\text{doubles on the third toss}\right)\\ P\left(\text{Doubles on the fourth toss}\right)\end{array}\right]\\ =0.833\times 0.833\times 0.833\times 0.167\\ =\text{0}\text{.0965}\end{array}$

Thus, the probability that first doubles occurs on the fourth toss is 0.0965.

The probability that first doubles occurs on the fifth toss is obtained below:

$\begin{array}{c}P\left(\text{Doubles on the fifth toss}\right)=\left[\begin{array}{l}P\left(\text{No}\text{doubles on the first toss}\right)\\ P\left(\text{No}\text{doubles on the second toss}\right)\\ P\left(\text{No}\text{doubles on the third toss}\right)\\ P\left(\text{No}\text{doubles on the fourth toss}\right)\\ P\left(\text{Doubles on the fifth toss}\right)\end{array}\right]\\ =0.833\times 0.833\times 0.833\times 0.833\times 0.167\\ =\text{0}\text{.0804}\end{array}$

Thus, the probability that first doubles occurs on the fourth toss is 0.0804.

The probability that first doubles occurs on the k^th toss is obtained below:

$\begin{array}{c}P\left(\text{Doubles on the }{k}^{\text{th}}\text{ toss}\right)=\left[\begin{array}{l}P\left(\text{No}\text{doubles on the first toss}\right)\\ P\left(\text{No}\text{doubles on the second toss}\right)\\ P\left(\text{No}\text{doubles on the third toss}\right)\\ P\left(\text{No}\text{doubles on the fourth toss}\right)\mathrm{...}\\ P\left(\text{Doubles on the }{k}^{\text{th}}\text{ toss}\right)\end{array}\right]\\ =\underset{\left(k-1\right)\text{times}}{\underbrace{0.833\times 0.833\times 0.833\times 0.833}}\times 0.167\\ ={\left(0.833\right)}^{k-1}\left(0.167\right)\end{array}$

(e)

To determine

To find: The probability to get to go again within three turns.

Answer to Problem 13.48E

The probability to get to go again within three turns is 0.4219.

Explanation of Solution

Calculation:

The probability that you get to go again within three turns is obtained below:

The required probability is as follows:

$P\left(\text{Go again within three turns}\right)=0.167+0.139+0.1159$

Thus, the probability to get to go again within three turns is 0.4219.