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Probability Practice Quest 2 Solution Guide

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Probability Practice Quest 2 Solution Guide . 1. Two cards are chosen at random from a deck of 52 cards without replacement. What is the probability of getting two kings? . 4/52 · 3/51 = 1/221. - PowerPoint PPT Presentation

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Page 1: Probability Practice Quest 2 Solution Guide

Probability Practice Quest 2 Solution Guide

Page 2: Probability Practice Quest 2 Solution Guide

1. Two cards are chosen at random from a deck of 52 cards without replacement. What is the probability of getting two kings? 4/52 · 3/51 = 1/221

2. In a shipment of 100 televisions, 6 are defective. If a person buys two televisions, what is the probability that both are defective if the first television is not replaced after it is purchased?

On a math test, 5 out of 20 students got an A. If three students are chosen at random without replacement, what is the probability that all three got an A on the test?

5/20 · 4/19 · 3/18 =1/114 Three cards are chosen at random from a deck of 52 cards without replacement. What is the probability of getting an ace, a king and a queen in order?

4/52 · 4/51 · 4/50 = 8/16575

A school survey found that 7 out of 30 students walk to school. If four students are selected at random without replacement, what is the probability that all four walk to school?

7/30 · 6/29 · 5/28 · 4/27 = 1/783

6/100 · 5/99 =1/330

Page 3: Probability Practice Quest 2 Solution Guide

At a middle school, 18% of all students play football and basketball, and 32% of all students play football. What is the probability that a student who plays football also plays basketball?

P( BB/ FB) = P(BB F B)/P(FB) = .18/32 = 9/16

In Europe, 88% of all households have a television. 51% of all households have a television and a VCR. What is the probability that a household with a television also has a VCR?

P(VCR/ TV) = P(VCR TV/P(TV) = 51/88

Page 4: Probability Practice Quest 2 Solution Guide

A research organization mailed out questionnaires to 10 random people. If the probability that any one person will answer the questionnaire is 10%, find the probability that at least 8 will answer

n = 10, k = 8, 9, 10, p = 10%, 1 - p = 90%

nCk pk(1-p)n-k

10C8 (.1)8(.9)2 +

10C9 (.1)9(.9)1 +

10C10 (.1)10

Page 5: Probability Practice Quest 2 Solution Guide

11)Of 200 students surveyed, 100 said they studied for their math test. 105 said they passed their test, and 80 said they studied and passed their test.

a. Incorporate the facts given above into a

conditional chart.P P

SS

Page 6: Probability Practice Quest 2 Solution Guide

11)Of 200 students surveyed, 100 said they studied for their math test. 105 said they passed their test, and 80 said they studied and passed their test.

a. Incorporate the facts given above into a

conditional chart.P P

S 80 100

S105 200

Page 7: Probability Practice Quest 2 Solution Guide

11)Of 200 students surveyed, 100 said they studied for their math test. 105 said they passed their test, and 80 said they studied and passed their test.

P PS 80 20 100

S 25 75 100105 95 200

P PS 80 100

S105 200

Page 8: Probability Practice Quest 2 Solution Guide

11)Of 200 students surveyed, 100 said they studied for their math test. 105 said they passed their test, and 80 said they studied and passed their test.

P PS 80 20 100

S 25 75 100105 95 200

a) P(pass)1052002140

Page 9: Probability Practice Quest 2 Solution Guide

11)Of 200 students surveyed, 100 said they studied for their math test. 105 said they passed their test, and 80 said they studied and passed their test.

P PS 80 20 100

S 25 75 100105 95 200

b) P(pass/studied) P(P/S) 80100 4 5

Page 10: Probability Practice Quest 2 Solution Guide

11)Of 200 students surveyed, 100 said they studied for their math test. 105 said they passed their test, and 80 said they studied and passed their test.

P PS 80 20 100

S 25 75 100105 95 200

_ _c) P(P/S) 75100 3 4

Page 11: Probability Practice Quest 2 Solution Guide

In a throw of a red die, r, and a white die, w, find:

· P (r = 5, given that r + w >

9) P (r = 5 | r + w > 9)

(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)

Page 12: Probability Practice Quest 2 Solution Guide

In a throw of a red die, r, and a white die, w, find:

· P (r = 5, given that r + w > 9)

P (r = 5 | r + w > 9) 3 10

(3, 6)

(4, 5) (4,6)

(5, 4) (5,5) (5,6)

(6, 3) (6,4) (6,5) (6,6)

Page 13: Probability Practice Quest 2 Solution Guide

In a throw of a red die, r, and a white die, w, find:

P (r + w = 9, given that

w = 6)

P (r + w = 9 |w = 6)

(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)

Page 14: Probability Practice Quest 2 Solution Guide

In a throw of a red die, r, and a white die, w, find:

P (r + w = 9, given that w = 6)

P (r + w = 9 |w = 6)

1 6

(1, 6)

(2, 6)

(3, 6)

(4, 6)

(5, 6)

(6, 6)

Page 15: Probability Practice Quest 2 Solution Guide

P(J K)

With Replacement 4 · 452 52 1 · 113 13 1_169

Without Replacement 4 · 4 52 51 1 · 413 51 4_663

Page 16: Probability Practice Quest 2 Solution Guide

There are 8 females and 5 males in my class. Half of the females are adults and 2 of the males are adults. If I randomly select 4 people (2 adults and 2 students) to go to the board, find P(all females). Leave in combination notation.

4C2 •2C0 •4C2 •3C0

13C4

Page 17: Probability Practice Quest 2 Solution Guide

(3x + 2y)4

1( )4 + 4( )3( )1+ 6( )2( )2+ 4( )1( )3+ 1( )4 1(3x)4 + 4(3x)3(2y)1+ 6(3x)2(2y)2+ 4(3x)1(2y)3+ 1(2y)4 1(81x4) + 4(27x3)(2y) + 6(9x2)(4y2)+ 4(3x)(8y3)+1(16y4) 81x4 + 216x3y) + 216x2y2 + 96xy3 + 16y4

Page 18: Probability Practice Quest 2 Solution Guide

fourth term of (2x – 3y)5

1( )5 + 5( )4( )1 + 10( )3( )2 + 10( )2( )3

10(2x)2(-3y)3

10(4x2)(-27y3)-1080x2y3

Page 19: Probability Practice Quest 2 Solution Guide

What is the probability of getting exactly 2 fives in 4 rolls of a die?

n = 4, k =2, P = 1/6, 1-P = 5/6

nCk pk(1-p)n-k

4C2(1/6)2(5/6)2 =6 (1/36)(25/36) =25/216

Page 20: Probability Practice Quest 2 Solution Guide

15)A dollar bill changer was tested with 100 one-dollar bills, of which 25 were counterfeit and the rest were legal. The changer rejected 30 bills, and 6 of the rejected bills were counterfeit. Construct a table, then find

a. Incorporate the facts given above into a

conditional chart.R R

CC

Page 21: Probability Practice Quest 2 Solution Guide

15)A dollar bill changer was tested with 100 one-dollar bills, of which 25 were counterfeit and the rest were legal. The changer rejected 30 bills, and 6 of the rejected bills were counterfeit. Construct a table, then find

a. Incorporate the facts given above into a

conditional chart.R R

C 6 25

C30 100

Page 22: Probability Practice Quest 2 Solution Guide

15)A dollar bill changer was tested with 100 one-dollar bills, of which 25 were counterfeit and the rest were legal. The changer rejected 30 bills, and 6 of the rejected bills were counterfeit. Construct a table, then find

R RC 6 19 25

C 24 51 7530 70 100

R RC 6 25

C30 100

Page 23: Probability Practice Quest 2 Solution Guide

15)A dollar bill changer was tested with 100 one-dollar bills, of which 25 were counterfeit and the rest were legal. The changer rejected 30 bills, and 6 of the rejected bills were counterfeit. Construct a table, then finda. The probability that a bill is legal

and it is accepted by the machine. P(CR)= 51 100b. The probability that a bill is

rejected, given it is legal. P(R/C)= 24 = 8 75 25

c. The probability that a counterfeit bill is not rejectedP(R/C) = 19 25

R RC 6 19 25

C 24 51 7530 70 100

Page 24: Probability Practice Quest 2 Solution Guide

A quiz has 6 multiple-choice questions, each with 4 choices. What is the probability of getting at least 5 of 6 questions correct?n = 6, k >5, P = .25, 1- P = .75

nCk pk(1-p)n-k

6C5(.25)5 (.75) + 6C6(.25)6

Page 25: Probability Practice Quest 2 Solution Guide

According to the National Institute of Health, 32% of all women will suffer a hip fracture due to osteoporosis by the age of 90. Six women aged 90 are randomly selected, find the probability that exactly 3 will have suffered a hip fractureat most 3 of them will havesuffered a hip fracturen = 6, k = < 3, p = 32%, 1 – p = 68%

nCk pk(1-p)n-k

6C3 (.32)3 (.68)3 + 6C2 (.32)2(.68)4 +

6C1 (.32)1 (.68)5 +

6C0 (.68)6

at least 1 of them will havesuffered a hip fracture.1 – 6C0 (.68)6