Question

A fair coin is tossed two ​times, and the events A and B are defined as shown below. Complete parts a through d.

​A: {At most one tail is​ observed}

​B: {The number of tails observed is even​}

a. Identify the sample points in the events​ A, B, Aunion​B, Upper A Superscript c​, and AintersectB.

Identify the sample points in the event A. Choose the correct answer below.

A. ​A:{TT comma HH​}

B. ​A:{TT​}

C. ​A:{HH comma HT comma TH​}

D. There are no sample points in the event.

Identify the sample points in the event B. Choose the correct answer below.

A. ​B:{TT​} B. ​

B:{TT comma HH​}

C. ​B:{HH comma HT comma TH​}

D. There are no sample points in the event.

Identify the sample points in the event AunionB. Choose the correct answer below.

A. Aunion​B:{TT​}

B. Aunion​B:{HH comma TH comma HT comma TT​}

C. Aunion​B:{TT comma HH​}

D. There are no sample points in the event.

Identify the sample points in the event Upper A Superscript c. Choose the correct answer below.

A. Upper A Superscript c​:{HH comma HT comma TH​}

B. Upper A Superscript c​:{HH​}

C. Upper A Superscript c​:{TT​}

D. There are no sample points in the event.

Identify the sample points in the event AintersectB. Choose the correct answer below.

A. Upper A intersect B: StartSet HH EndSet

B. Aintersect​B:{TT​}

C. Aintersect​B:{HH comma HT comma TH​}

D. There are no sample points in the event.

Find​ P(A), P(B), ​P(Aunion​B),

Upper P left parenthesis Upper A Superscript c Baseline right parenthesis​, and ​P(Aintersect​B) by summing the probabilities of the appropriate sample points. ​P(A)equals 3/4 three fourths ​(Type an integer or simplified​ fraction.) Find​ P(B). ​P(B)equals 1/4 one fourth ​(Type an integer or simplified​ fraction.) Find ​P(Aunion​B). ​P(Aunion​B)equals 1 1 ​(Type an integer or simplified​ fraction.) Find Upper P left parenthesis Upper A Superscript c Baseline right parenthesis. Upper P left parenthesis Upper A Superscript c Baseline right parenthesisequals nothing ​(Type an integer or simplified​ fraction.) Find ​P(Aintersect​B). ​P(Aintersect​B)equals 0 0 ​(Type an integer or simplified​ fraction.) c. Find ​P(Aunion​B) using the additive rule. Compare your answer to the one you obtained in part b. ​P(Aunion​B)equals nothing ​(Type an integer or simplified​ fraction.) Compare your answer to the one you obtained in part b. Choose the correct answer below. A. The value of ​P(Aunion​B) calculated using the additive rule is less than ​P(Aunion​B) calculated by summing the probabilities of the sample points. B. ​P(Aunion​B) calculated using the additive rule is greater than ​P(Aunion​B) calculated by summing the probabilities of the sample points. C. Both calculations of ​P(Aunion​B) produce the same result. d. Are events A and B mutually​ exclusive? Why? A. No​, events A and B are not mutually exclusive because ​P(Aintersect​B)equals0. B. Yes​, events A and B are mutually exclusive because ​P(Aintersect​B)equals0. C. No​, events A and B are not mutually exclusive because ​P(Aintersect​B)not equals0. D. Yes​, events A and B are mutually exclusive because ​P(Aintersect​B)not equals0.

Answer 1

Solution :

Fair coin is tossed two times. H : getting Head, T: getting tail

Sample space = S = {HH, HT, TH, TT}

Event A & B are defined as below

A : At most one tail is observed = {HH, HT, TH}

B : the number of tails is observed are even = {TT}

Ac = {TT}

AUB = {HH, HT, TH, TT}

A∩B = Φ = Null/Empty set

1.

a)

Identify the sample points in event A   

Option C is correct. - A: {HH, HT, TH}

b)

Identify the sample points in event B.

Option A is correct   - B: {TT}

c)

Identify the sample points in event AUBc

Option B is correct - AUB: {HH, HT, TH, TT}

d)

Identify the sample points in event Ac

Option C is correct - Ac: {TT}

e)

· Identify the sample points in event A∩B

Option D is correct - There are no sample points in the event

2.

Probability of sample point = total sample point in event / Total sample space

· P(A) = ¾ =0.75

· P(B) = ¼ = 0.25

· P(AUB) = 4/4 =1

· P(Ac) = ¼ = 0.25

· P(A∩B) = 0/4 = 0

3.

find P(AUB) using additive rule

P(AUB) = P(A) + P(B) – P(A∩B)

=0.75 + 0.25 – 0
= 1

P(AUB) = 1

Option C is correct -  Both calculations of P(AUB) produce the same results

4.

Are events A & B are mutually exclusive? Why?

Option B is correct - Yes. Events A & B are mutually exclusive because P(A∩B) = 0

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