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A pair of fair dice is tossed. Define the following events:

A: [Exactly one of the dice shows a 1.]

B: [The sum of the numbers on the two dice is even.]

a. Identify the sample points in the events A,B,A∩B,A∪B,andAc.

b. Find the probabilities of all the events from part a by summing the probabilities of the appropriate sample points.

C. Using your result from part b, explain why A and B are not mutually exclusive.

d. Find P(A∪B) using the additive rule. Is your answer the same as in part b?

Short Answer

Expert verified
  1. A=[(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(3,1)(4,1)(5,1)(6,1)]

B=[(1,1)(1,3)(1,5)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(4,2)(4,4)(4,6)(5,1)(5,3)(5,5)(6,2)(6,4)(6,6)]

A∩B=[(1,1)(1,3)(1,5)(3,1)(5,1)]

A∪B=[(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(4,1)(4,2)(4,4)(4,6)(5,1)(5,3)(5,5)(6,1)(6,2)(6,4)(6,6)]

Ac=[(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)(3,3)(3,4)(3,5)(3,6)(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,3)(5,4)(5,5)(5,6)(6,2)(6,3)(6,4)(6,5)(6,6)]

b.P(A)=1136,P(B)=1836,P(A∩B)=536,P(A∪B)=2436,P(Ac)=2536

c. No

d. Yes

Step by step solution

01

Identify the sample points

The sample points of the sample space are the primary outcomes of an experiment. Sample points are sample space elements that represent the experiment in terms of the sample space. Sample points are often referred to as sampling units or observations.

Rolling a pair of fair dice gives the following sample points:

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

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

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

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

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

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

02

Identify the sample points

A,B,A∩B,A∪B,andAc.

A=[(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(3,1)(4,1)(5,1)(6,1)]

B=[(1,1)(1,3)(1,5)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(4,2)(4,4)(4,6)(5,1)(5,3)(5,5)(6,2)(6,4)(6,6)]

A∩B=[(1,1)(1,3)(1,5)(3,1)(5,1)]

A∪B=[(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(4,1)(4,2)(4,4)(4,6)(5,1)(5,3)(5,5)(6,1)(6,2)(6,4)(6,6)]

Ac=[(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)(3,3)(3,4)(3,5)(3,6)(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,3)(5,4)(5,5)(5,6)(6,2)(6,3)(6,4)(6,5)(6,6)]

03

Calculation of the probabilities of all the events from part a summing the probability of the data from a sample point

Here, the total number of sample points is 36.

Likewise,

n=36n(A)=11n(B)=18n(A∩B)=5

n(A∪B)=24n(Ac)=25

Now, summing the probabilities of the data from a sample point is:

P(A)=1136,P(B)=1836,P(A∩B)=536,P(A∪B)=2436,

P(Ac)=2536

04

Identify why A and B are not mutually exclusive.

IfP(A∩B)=0, A and B are hence mutually exclusive.

Here,

P(A∩B)=536≠0

Hence, A and B don't get to be mutually exclusive.

05

Find and identify the answer is the same as in part b.

P(A∪B)=P(A)+P(B)−P(A∩B)=1136+1836−536=11+18−536=2436

Yes, it is the same as in part b.

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