CHAPTER 11
Modeling Randomness Section 11.1
11.1.13
11.1.1 C.
a. As the number of dice increases, the probability of getting at least one 5 or 6 will also increase. (Imagine rolling 100 dice. It is almost a sure thing that you will get at least one 5 or 6.) Therefore the probability that the price will be 50 cents or less will decrease as the number of dice increases. Also, rolling additional dice will never drop a price from above 50 cents to below it, but it can move it from below 50 cents to above it.
11.1.2 B. 11.1.3 C. 11.1.4 E. 11.1.5 A. 11.1.6 a. Yes b. Yes c. No d. No e. Yes f. Yes (but not too much longer using a computer) 11.1.7 Number the index cards 1, 2, and 3 (they each represent a phone). Shuffle them and place them on the table in spots labeled 1, 2, and 3. If all the numbers on the cards fail to match all the numbered spots on the table call that a success; anything else call a failure. Repeat this 1,000 times. The number of successes divided by 1,000 is an estimate for the probability none of them would get their own phones. 11.1.8 a. ABC, ACB, BAC, BCA, CAB, CBA b. BCA, CAB
b. Because getting a “large” number is more likely with the more dice, the price is more likely to be higher and so the mean price will also increase as the number of dice increases. c. If you can choose one die (and you don’t have to multiply the outcome by 10) then you will always be able to afford the ice cream cone! However, if you have to use more than one die, you should choose two dice because the probability of an outcome of 50 cents or less is the greatest. 11.1.14 a. 1/6(1/6)(2) + 1/6(2/6)(4) = 10/36 b. (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) P(consecutive) = 10/30 c. The cards. This makes sense because while the numerator is the same, six pairs were eliminated from the denominator when we switched to the cards, therefore that probability increased in value.
c. 2/6 = 1/3
11.1.15
d. If the three executives dropped their phones and randomly picked them up many, many times, about 1/3 of the time nobody would get their own phone.
a. B1B2B3, G1B2B3, B1G2B3, B1B2G3, G1G2B3, B1G2G3, G1B2G3, G1G2G3 b. B1B2B3, G1B2B3, B1G2B3, B1B2G3
11.1.9
11.1.16
a. 1/2 b. 0 c. 1/6 d. 2/3 11.1.10
c. 2/8 = 1/4 a. B1B2B3B4, G1B2B3B4, B1G2B3B4, B1B2G3B4, B1B2B3G4, G1G2B3B4, G1B2G3B4, G1B2B3G4, B1G2G3B4, B1G2B3G4, B1B2G3G4, B1G2G3G4, G1B2G3G4, G1G2B3G4, G1G2G3B4, G1G2G3G4 b. B1B2B3B4, G1B2B3B4, B1G2B3B4, B1B2G3B4, B1B2B3G4 c. 2/16 = 1/8 d. A 3 to 1 breakdown is more likely because this probability is 8/16 whereas two boys and two girls is 6/16. 11.1.17
11.1.11
11.1.12
a. Write “Male” on four cards and “Female” on two cards. Shuffle the six cards and randomly choose two of them. If they are the two Female cards call this a success. Repeat this for a total of 1,000 times. The proportion of successes out of 1,000 is an estimate for the probability. b. If you do it with dice, you can think each number represents a different person. However, you could roll doubles (e.g. (3,3)) and
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