The geometric distributions You are tossing a pair of fair, six-sided dice in a board game. Tosses are independent. You land in a danger zone that

requires you to roll doubles (both faces showing the same number of spots) before you are allowed to play again. How long will you wait to play again?

(a) What is the probability of rolling doubles on a single toss of the dice? (If you need review, the possible outcomes appear in Figure 5.2 (page 300). All $36$ outcomes are equally likely.)

(b) What is the probability that you do not roll doubles on the first toss, but you do on the second toss?

(c) What is the probability that the first two tosses are not doubles and the third toss is doubles? This is the probability that the first doubles occurs on the

third toss.

(d) Now you see the pattern. What is the probability that the first doubles occurs on the fourth toss? On the fifth toss? Give the general result: what is the probability that the first doubles occurs on the kth toss?

In a toss, the number of dice rolled equals$2$

The number of sides of a die is $6$

$Probability=\frac{Numberoffavorableoutcomes}{Totalnumberofoutcomes}$

Complementary rule: $P(notA)=1-P\left(A\right)$

On rolling a pair of dice, the sample space is shown below:

Number of outcomes in total $6^6=36$

The results of doubles=$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$

As a result, the likelihood of rolling doubles is$\frac{6}{6^2}=\frac{1}{6}=0.1667$

From the sub part a, $1/6$chance of rolling doubles .

The likelihood of not rolling dice is based on the complements rule, as explained above$1-\frac{1}{6}=\frac{5}{6}$

As a result, the necessary probability$\frac{5}{6}\times \frac{1}{6}=\frac{5}{6}=0.13889$

From the above sub parts $$\begin{gathered} P\left(double\right)=1/6 \\ P(notdouble)=5/6 \end{gathered}$$

As a result, the chances of a double on the third throw but not the first or second will be higher$\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=0.1157$

As a result, the likelihood is $0.1157$

Let$X$ be the number of tosses required for a double.

Doubles are more likely to occur in the fourth toss.

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

Probability of a double in the fifty-fifth toss

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

It can be observed that the result is increased by $5/6$ each time.

$P(X=4)=P(X=3)\times \frac{5}{6}=\frac{5^2}{6^3}\times \frac{5}{6}=\frac{5^3}{6^4}$

This is consistent with the general rule.

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