Question

Craps. In the game of craps, a player rolls two balanced dice. thirty six equally likely outcome are possible.

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.

Part (a) Compute the probability of each of the seven events.

Part (b) The player wins on the first roll if the sum of the dice 2,3 or 12. Find the probability of that event by using the special addition rule and your answer from part (a).

Part (c) The player loses on the first roll if the sum of the dice 7 or 11. Find the probability of that event by using the special addition rule and your answer from part (a).

Part (d) Compute the probability that either the sum of the dice is 8 or double are rolled without using the general addition rule.

Part (e) Compute the probability that either the sum of the dice is 8 or double are rolled without using the general addition rule. and compare your answer to the you obtained in part (d)

Short answer

Part(a)

A = 0.167

B = 0.056

C= 0.028

D = 0.056

E = 0.028

F = 0.139

G = 0.167

Part (b) $P(A\cup B)=0.223$

Part (c) $P(C\cup D\cup B)=0.112$

Part (d) $P(F\cup G)=0.278$

Part (e) From the results of part (d) and part (e), observe that the obtained answers are same.

$P(F\cup G)=0.278$

Step-by-step solution

Part (a) Step 1. Given information.

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.

Part (a) Step 2.  Compute the probability of each the seven events.

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,

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

Part (b) Step 1. Determine whether the sum of the dice is 7 or 11.

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,

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

As a result, 0.223 is the required probability.

Part (c) Step 1. Determine whether the sum of the dice is 2,3 or 12.

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,

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

As a result, 0.112 is the required probability.

Part (d) Step 1. probability that the sum of the dice will be 8 or that doubles will be rolled.

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

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

Part (e) Step 1. probability that the sum of the dice will be 8 or that doubles will be rolled.

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:

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

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

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