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# Conditional Probability of Gambling

1.4-13 Tn 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 rule is, If the bettor rolls a 7 before the "point," the bettor loses; hut if the "point" is rolled before a 7, the bettor wins.
(a) List the 36 outcomes in the sample space for the roll of a pair of dice. Assume that each of them has a probability of 1/36.
(b) Find the probability that the bettor wins on the first roll. That is, fInd the probability of rolling a 7 or ii, P(7 or 11).
(c) Given that 8 is the outcome on the first roll, find the probability that the bettor now rolls the point 8 before rolling a 7 and thus wins. Note that at this stage in the game the only outcomes of interest are 7 and 8. Thus find P(8 | 7 or 8).
(d) The probability that a bettor rolls an 8 on the first roll and then wins is given by P(8)P(8 | 7 or 8). Show that this probability is (5/36)(5/11).
(e) Show that the total probability that a bettor wins in the game of craps is 0.49293.
HINT: Note that the bettor can win in one of several mutually exclusive ways: by rolling a 7 or ii on the first roll or by establishing one of the points 4, 5, 6. 8, 9, or 10 on the first roll and then obtaining that point before a 7 on successive rolls.

#### Solution Summary

Conditional probability is investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

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