Question 547845: Two dice are tossed and various amounts are paid according to the outcome. In a certain game if a nine or ten occurs on the first roll the players wins. what is the probability of winning on the first roll.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! There are 36 combinations (six outcomes on the first die times 6 outcomes on the second die) that can be rolled when you roll a two dice.
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How many different combinations that total to 9 can be rolled? (First die 6, second die 3); (First die 3, second die 6); (First die 5, second die 4); (First die 4, second die 5). So there are 4 possible combinations totaling 9.
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How many different combinations that total to 10 can be rolled? (First die 6, second die 4); (First die 4, second die 6); (Both dies 5). So there are 3 possible combinations totaling 9.
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This means that there are a total of 7 chances of winning (4 from a 9 and 3 from a 10) by scoring a 9 or a 10 on the first roll.
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Therefore, the chance of winning on the first roll is 7 in 36 or the probability is 7 divided by 36 and this is 0.1944 which is expressed as a probability of 19.44%.
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Hope this helps you to understand the problem. And to help you with future problems involving a dice pairs, here are the possibilities for each roll:
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Rolling a 2: one chance (1&1)
Rolling a 3: two chances (2&1)(1&2)
Rolling a 4: three chances (3&1)(1&3)(2&2)
Rolling a 5: four chances (4&1)(1&4)(3&2)(2&3)
Rolling a 6: five chances (5&1)(1&5)(4&2)(2&4)(2&2)
Rolling a 7: six chances (6&1)(1&6)(5&2)(2&5)(4&3)(3&4)
Rolling an 8: five chances (6&2)(2&5)(5&3)(3&5)(4&4)
Rolling a 9: four chances (6&3)(3&6)(5&4)(4&5)
Rolling a 10: three chances (6&4)(4&6)(5&5)
Rolling an 11: two chances (6&5)(5&6)
Rolling a 12: one chance (6,6)
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The sum of the numbers of possible outcomes is: 1+2+3+4+5+6+5+4+3+2+1 = 36
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