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Probability Practice Quest 2 Solution Guide

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.

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Probability Practice Quest 2 Solution Guide

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  1. Probability Practice Quest 2 Solution Guide

  2. 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? 6/100 · 5/99 =1/330 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

  3. 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

  4. 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% nCkpk(1-p)n-k 10C8 (.1)8(.9)2 + 10C9 (.1)9(.9)1 + 10C10 (.1)10

  5. 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.

  6. 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.

  7. 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.

  8. 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(pass) 105 200 21 40

  9. 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. b) P(pass/studied) P(P/S) 80 100 4 5

  10. 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. _ _ c) P(P/S) 75 100 3 4

  11. 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)

  12. 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

  13. 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)

  14. 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

  15. P(J  K) With Replacement 4 4 • 52 1  1 13 13 1_ 169 Without Replacement 4 4 52 51 1  4 13 51 4_ 663

  16. 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

  17. (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

  18. 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

  19. 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 nCkpk(1-p)n-k 4C2(1/6)2(5/6)2 = 6 (1/36)(25/36) = 25/216

  20. 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.

  21. 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.

  22. 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

  23. 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 • The probability that a bill is legal and it is accepted by the machine. P(CR)= 51 100 • The probability that a bill is rejected, given it is legal. P(R/C)= 24 = 8 75 25 • The probability that a counterfeit bill is not rejected P(R/C) = 19 25

  24. A quiz has 6 multiple-choice questions, each with 4 choices. What is the probability of getting at least5 of 6 questions correct? n = 6, k >5, P = .25, 1- P = .75 nCkpk(1-p)n-k 6C5(.25)5 (.75) + 6C6(.25)6

  25. 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 fracture at most 3 of them will have suffered a hip fracture n = 6, k = < 3, p = 32%, 1 – p = 68% nCkpk(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 have suffered a hip fracture. 1 – 6C0 (.68)6

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