91Ó°ÊÓ

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$\frac{5}{6}×\frac{1}{6}=\frac{5}{6}=0.13889$

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$\frac{5}{6}×\frac{5}{6}×\frac{1}{6}=0.1157$

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.

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

Probability of a double in the fifty-fifth toss

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

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

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

This is consistent with the general rule.

$P(X=k)=\frac{5^{k−1}}{6^k}$