S. In the gambling game craps, a pair of dice is rolled and the outcome of the e
ID: 3069236 • Letter: S
Question
S. In the gambling game craps, a pair of dice is rolled and the outcome of the experiment is the sum of the points on the up-sides of the six-sided dice. The bettor wins on the first roll if the sum is 7 or 11. The bettor loses on the first roll if the sum is 2,3, or 12.If the sum is 4,5,6,8,9, or 10, that number is called the bettor's "point": Once the "point" is established, the following rule is used: If the bettor rolls a 7 before rolling the "point" again, he loses. But if he rolls the "point" again before rolling a 7, he wins. (4 points) There are at least two ways of thinking of the sample space here: either S 2,3,4,5,6.7,8,9,10,11,12) or S a. 1.1).(1.2),. 1.6),(2.1),(2.2)... 2,6),3,.1)..(6,6) (the first represents the sum of the numbers on the dice while the second gives the numbers rolled on each die). Why is it easier to use the second S when calculating probabilities? b. Find the probability that the bettor wins on the first roll. Given that 8 is the outcome on the first roll,find the probability that the bettor wins Hint: We only care about two passible outcomes, rolling a 7 or an 8. With that fact, you should be able to calculate the conditional probability that is being asked for here. c. d. Find the probability that the bettor rolls an 8 on the first roll and goes on to win. 6. Continuing with the craps example, show that the total probability that the bettor wins a game of craps is 0.49293. (3 points) 7. In the NCAA basketball tournament, 64 teams compete for the championship (really, it's 68, but we'll ignore the 4 "play-in" teams). Below is the bracket at the opening round for the 2018 men's tournament if you are unfamiliar with this tournament. It is popular for people to participate in competitions where brackets are filled out before the tournament. Prizes are often given after the tournament to those whose predictions most closely match what actually happened. (3 points) It is often claimed that the probability of filling out a perfect bracket is one out of 2.9 quintillion (1/9,223,372,036,854,775,808). Show how this probability is calculated. a. b. What assumptions are being made to come up with one out of 9.2 quintillion? c Many mathematicians and statisticians claim that this assumption is not reasonable and the probability of filing out a perfect bracket is actually larger than one out of 9.2 quintillion (i.e. it's more likely to happen). What are some possible factors that would invalidate the assumptions you listed in b? How would these factors increase the probability of filling out a perfect bracket?Explanation / Answer
Answer to question# 5)
Answer to part a)
The second set makes it clear about all the possible outcomes and it is easier to calculate the probability of each case
If two dice are rolled, there are supposed to be 36 pairs. Knowing them individually and then finding the sum provides an easier way to note the probability of each sum
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Answer to part b)
The bettor would win in the first roll if he gets a sum of 7 or 11
Number of outcomes with sum 7 = (3,4) (4,3) (2,5) (5,2)
Number of outcomes with sum 11 = (4,5) (5,4)
Thus total number of winning outcomes = 6
Total outcomes = 36
P (win) = 6/36 = 1/6
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Answer to part c)
P(8) = (5,3) (3,5) (4,4) (6,2) (2,6) total 5 events
So P(8) = 5/36
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P(win) = 1/6
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P(win | 8) = P(win AND 8) / P(8)
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Since the outcome on second roll is independent on the outcome in the first roll
P(Win AND 8) = P(win) * P(8)
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Thus P(win | 8) = P(win)*P(8) / P(8)
P(win | 8) = P(win)
P(win | 8) = 1/6
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Answer to part d)
P(8 AND win) = P(8) * P(win)
P(8 AND win) = 5/36*6/36 = 5 / 216
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