91Ó°ÊÓ

A = event the sum of the dice is 7

B = event the sum of the dice is 11

C = event the sum of the dice is 2

D = event the sum of the dice is 3

E = event the sum of the dice is 12

F = event the sum of the dice is 8

G = event double are rolled.

Let A be the occurrence in which the sum of the dice equals 7.

(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) is the sample space for event.

There are six favourable scenarios for occurrence A.

The likelihood of incident A is,

$$\begin{gathered} P\left(A\right)=\frac{P\left(A\right)}{N} \\ P\left(A\right)=\frac{6}{36} \\ P\left(A\right)=\frac{1}{6} \\ P\left(A\right)=0.167 \end{gathered}$$

Let B be the occurrence in which the sum of the dice equals 11.

(5,6),(6,5) is the sample space for event.

There are two favourable scenarios for occurrence B.

The likelihood of incident B is

$$\begin{gathered} P\left(B\right)=\frac{P\left(B\right)}{N} \\ P\left(B\right)=\frac{2}{36} \\ P\left(B\right)=\frac{1}{18} \\ P\left(B\right)=0.056 \end{gathered}$$

Let C be the occurrence in which the sum of the dice equals 2.

(1,1) is the sample space for event.

There are one favourable scenarios for occurrence C.

The likelihood of incident C is

$$\begin{gathered} P\left(C\right)=\frac{P\left(C\right)}{N} \\ P\left(C\right)=\frac{1}{36} \\ P\left(C\right)=0.028 \end{gathered}$$

Let D be the occurrence in which the sum of the dice equals 3.

(1,2),(2,1) is the sample space for event.

There are two favourable scenarios for occurrence D.

The likelihood of incident D is

$$\begin{gathered} P\left(D\right)=\frac{P\left(D\right)}{N} \\ P\left(D\right)=\frac{2}{36} \\ P\left(D\right)=0.056 \end{gathered}$$

Let E be the occurrence in which the sum of the dice equals 12.

(6,6) is the sample space for event.

There are one favourable scenarios for occurrence E.

The likelihood of incident E is

$$\begin{gathered} P\left(E\right)=\frac{P\left(E\right)}{N} \\ P\left(E\right)=\frac{1}{36} \\ P\left(E\right)=0.028 \end{gathered}$$

Let F be the occurrence in which the sum of the dice equals 8.

(6,2), (2,6), (4,4),(5,3),(3,5) is the sample space for event.

There are one favourable scenarios for occurrence F.

The likelihood of incident F is

$$\begin{gathered} P\left(F\right)=\frac{P\left(F\right)}{N} \\ P\left(F\right)=\frac{5}{36} \\ P\left(F\right)=0.139 \end{gathered}$$

Let G be the event of the double roll dice = 8.

The sample space for event G is {(1,l),(2,2).(3,3),(4,4),(S,S),(6,6)}.

The number of favorable cases for event G is 5.

The probability of event G is,

$$\begin{gathered} P\left(G\right)=\frac{P\left(G\right)}{N} \\ P\left(G\right)=\frac{6}{36} \\ P\left(G\right)=0.167 \end{gathered}$$

The sum of the dice has a probability of 7 = 0.167.

The sum of the dice has a probability of 11 = 0.056.

The probability of the sum of the dice being 7 or 11 are,

$$\begin{gathered} P(A∪B)=P\left(A\right)+P\left(B\right) \\ P(A∪B)=0.167+0.056 \\ P(A∪B)=0.223 \end{gathered}$$

As a result, 0.223 is the required probability.

The sum of the dice has a probability of 2 = 0.028.

The sum of the dice has a probability of 3 = 0.056.

The sum of the dice has a probability of 11 = 0.056.

The probability of the sum of the dice being 2, 3 or 11 are,

$$\begin{gathered} P(C∪D∪B)=P\left(C\right)+P\left(D\right)+P\left(B\right) \\ P(C∪D∪B)=0.028+0.056+0.028 \\ P(C∪D∪B)=0.112 \end{gathered}$$

As a result, 0.112 is the required probability.

Let F be the occurrence in which the sum of the dice equals 8.

Assume that G is the event in which doubles are rolled.

For event F or G, the sample space is

(2,6),(3,5),(6,2),(5,3),(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)

There are ten cases that are favourable.

If the total of the dice is 8, or if doubles are thrown, the probability is

$$\begin{gathered} P(F∪G)=\frac{N(F∪G)}{N} \\ P(F∪G)=\frac{10}{36} \\ P(F∪G)=0.278 \end{gathered}$$

Let F be the occurrence in which the sum of the dice equals 8.

Assume that G is the event in which doubles are rolled.

The following are the probability values:

$$\begin{gathered} P\left(F\right)=\frac{5}{36} \\ P\left(G\right)=\frac{1}{36} \\ P(Fâˆ\copyright G)=\frac{1}{36} \end{gathered}$$

The probability of rolling an 8 of doubles on either side of the dice is

$$\begin{gathered} P(F∪G)=P\left(F\right)+P\left(G\right)-P(Fâˆ\copyright G) \\ P(F∪G)=\frac{5}{36}+\frac{1}{6}-\frac{1}{36} \\ P(F∪G)=\frac{10}{36} \\ P(F∪G)=0.278 \end{gathered}$$