Jim plays crap he knows if he hit 7 than 11 he could win a considerable money. He decides to gamble all his money for that possibility. A) what is the probability of hitting those #'s?
I presume you mean Jim plays "Craps", not Jim plays crap! And "7 then 11" rather than "7 than 11". With those assumptions...
In craps, two dice are rolled and players bet on the outcome (the sum of the two dice). If Jim requires:
1. The first roll to be a 7, AND
2. The second roll to be 11,
then we can simply multiply the probability of each outcome to find the probability of BOTH occurring (and Jim winning his jackpot).
There are 6 combinations of 2 dice that sum to 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), and 36 combinations in total, so the probability of getting a 7 on the first roll is:
P(7) = 6/36 = 1/6
There are just two combinations that sum to 11 though (5+6 or 6+5):
P(11) = 2/36 = 1/18
So the probability of both occurring is:
P(7 then 11) = P(7) x P(11)
= 1/6 x 1/18
So Jim has a 1-in-108 chance (a probability of 0.926%).
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