Chapter 5.3, Problem 102E

(a)

To determine

To find: Probability of rolling doubles on the second toss but not on the first.

Answer to Problem 102E

The probability is 0.1667.

Explanation of Solution

Given:

Number of dices rolled in a toss = 2

Number of sides in a die = 6

Formula used:

The probability can be calculated with the formula mentioned below

  $\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

The rule of complements states

  $P\left(\text{not A}\right)=1-P\left(A\right)$

Calculation:

The sample space on rolling a pair of dice is

	1	2	3	4	5	6
1	1,1,	1,2	1,3	1,4	1,5	1,6
2	2,1	2,2	2,3	2,4	2,5	2,6
3	3,1	3,2	3,3	3,4	3,5	3,6
4	4,1	4,2	4,3	4,4	4,5	4,6
5	5,1	5,2	5,3	5,4	5,5	5,6
6	6,1	6,2	6,3	6,4	6,5	6,6

Total number of outcomes $6^2=36$

The outcomes with doubles $=\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)$

So, the probability of rolling doubles is $\frac{6}{6^2}=\frac{1}{6}=0.1667$

(b)

To determine

To find: Probability of rolling doubles on the second toss but not on the first.

Answer to Problem 102E

The probability is $0.1389$

Explanation of Solution

Calculation:

From the sub part a,

Probability of rolling doubles $=\frac{1}{6}$

Using complements rule as mentioned above, the probability of not rolling dice is $1-\frac{1}{6}=\frac{5}{6}$

Therefore, the required probability $\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}=0.13889$

(c)

To determine

To find: The probability of first double occurring in the third toss.

Answer to Problem 102E

The required probability is $0.1157$

Explanation of Solution

Calculation:

From the above sub parts

  $\begin{array}{l}P\left(\text{double}\right)=\frac{1}{6}\\ P\left(\text{not double}\right)=\frac{5}{6}\end{array}$

So, the probability of a double occurring on the third toss but not on the first and second will be

  $\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{5^2}{6^3}=0.1157$

Thus, the probability is 0.1157.

(d)

To determine

To find: The probability of occurring doubles on the kth toss.

Answer to Problem 102E

  $P\left(X=k\right)=\frac{5^{k-1}}{6^k}$

Explanation of Solution

Let X be the number of tosses for doubles to occur

Probability of doubles occurring in the fourth toss

  $P\left(X=4\right)=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{5^3}{6^4}$

Probability of doubles occurring in the fifth toss

  $P\left(X=5\right)=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{5^4}{6^5}$

It can be seen that each time the result is being multiplied by $\frac{5}{6}$

  $\begin{array}{l}P\left(X=4\right)=P\left(X=3\right)\times \frac{5}{6}\\ =\frac{5^2}{6^3}\times \frac{5}{6}=\frac{5^3}{6^4}\end{array}$

This follows the general rule

  $P\left(X=k\right)=\frac{5^{k-1}}{6^k}$