6. Casinos rely on the laws of probability and expected a will walk away very w

Question

6. Casinos rely on the laws of probability and expected a will walk away very w of CRAPS guarantee them profits on a daily basis. Some individva take a look at the gaie while others will leave with nothing but memories Ler individuals will walk away very wealthy the sum is any other t the sum is 7 or 11, you win. If the sum is 2, 3, or 12, you r, you roll again. In fact, you continue throwing the Roll a pair of six-sided dice. If 7 or 11 dice until you either roll that same number egain (WINI) or roll a 7 (LOSE). What is the probability that you obtain a sum of 7 or a sum (Remember the sample space for rolling 2 dice.) a. of 11 on the first roll? b. What is the probability that you obtain a sum of 2, 3, or 12 on the first roll? c. What is the probability you will have to roll again?th (You got something other than a 7, 11,2, 3, or d. The probability of winning CRAPS is 0.493 (based on the calculations above and geometric probability of "waiting" until the same number appears.) 0.493 for winning one game of CRAPS, suppose you feel ad Find each probability using the rules learned in this chapter Knowing the probability is adventurous, and play 2 games What is the probability that you win both games? (Be careful, your probability should not be getting larger..... i. i. What is probability you win the first game and lose the second? ili. What is the probability you lose both games? iv. Why do casinos make money off of this game???

Explanation / Answer

possible outcome for sum of 7 and 11 = 8

total possible outcome = 6^2 =36

so, P( sum of 7 or sum of 11 on first roll) = 8/36 = 0.2222

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b)

possible outcome for sum of 2 ,3 or 12=4

so, P(sum of 2, 3, or 12 on first roll) = 4/36 =0.1111

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c)

P(rolling again) = 1 - P(sum of 7,11) - P(sum of 2,3 or 12)

=1-0.2222-0.1111

=0.6667

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d1)

P(winning) = 0.493

both games are indepenent of each other ,so

P(winning both games) = 0.493*0.493

=0.243

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d2)

P(winning first game and losing second) = 0.496*(1-0.493)

=0.24995

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d3)

P(lose both games) = (1-0.493)*(1-0.493)

=0.2570

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1,6 5,6 2,5 6,5 3,4 4,3 5,2 6,1

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