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A certain dice game can only be won if, when a player throws [#permalink]
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22 Jul 2016, 08:22
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A certain dice game can only be won if, when a player throws two fair sixsided dice, the number showing on one of the dice is a multiple of the number showing on the other. What is the probability that a player wins this game? (A) 11/18 (B) 11/36 (C) 1/4 (D) 2/9 (E) 7/36 My approach is as follows : For first dice, with 1, there will be 6 ways, Similarly for 2 on first dice, there will be 3 ways (2,4,6) For 3, on first dice, there will be 2 ways (3,6) For 4, 5 ,6 , on first dice, 1 way for each. The answer doesn't seem to be right. Am I missing out on any possibility? (
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Re: A certain dice game can only be won if, when a player throws [#permalink]
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22 Jul 2016, 08:55
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aditi2013 wrote: [My approach is as follows : For first dice, with 1, there will be 6 ways, Similarly for 2 on first dice, there will be 3 ways (2,4,6) For 3, on first dice, there will be 2 ways (3,6) For 4, 5 ,6 , on first dice, 1 way for each. The answer doesn't seem to be right. Am I missing out on any possibility? First dice with 1 > 6 ways First dice with 2 > 3 ways First dice with 3 > 2 ways First dice with 4, 5 and 6 > 1 + 1 + 1 = 3 ways Total = 14 Now the entire procedure can be repeated for second dice as well > 14 ways We have double counted (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6) > 6 Required number of ways = 14 + 14  6 = 22 ways Probability = 22/(6 * 6) = 22/36 = 11/18 Answer: A



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Re: A certain dice game can only be won if, when a player throws [#permalink]
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22 Jul 2016, 10:21
Vyshak wrote: aditi2013 wrote: [My approach is as follows : For first dice, with 1, there will be 6 ways, Similarly for 2 on first dice, there will be 3 ways (2,4,6) For 3, on first dice, there will be 2 ways (3,6) For 4, 5 ,6 , on first dice, 1 way for each. The answer doesn't seem to be right. Am I missing out on any possibility? First dice with 1 > 6 ways First dice with 2 > 3 ways First dice with 3 > 2 ways First dice with 4, 5 and 6 > 1 + 1 + 1 = 3 ways Total = 14 Now the entire procedure can be repeated for second dice as well > 14 ways We have double counted (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6) > 6 Required number of ways = 14 + 14  6 = 22 ways Probability = 22/(6 * 6) = 22/36 = 11/18 Answer: A Thank you Vyshak. Its pretty clear now.



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Re: A certain dice game can only be won if, when a player throws [#permalink]
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23 Jul 2017, 11:15
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Or simply just list what the question asks, i.e. pairs in which one is the factor of another: 1pairs with (1,2,3,4,5,6) 2pairs with (1,2,4,6) 3pairs with (1,3,6) 4pairs with (1,2,4) 5pairs with (1,5) 6pairs with (1,2,3,6)
total 22 such pairs. p=22/36=11/18.



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Re: A certain dice game can only be won if, when a player throws [#permalink]
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11 Sep 2017, 22:43
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D1d2 11,2,3,4,5,6 21,2,4,6 31,3,6 41,2,4 51,5 61,2,3,6 DICE 1 ON THE LEFT FOLLOWED BY DASH AND THEN DICE 2 OUTCOMES FOR WIN TOTAL = 22 CASES POSSIBLE TOTAL = 6X6=36 P = 22/36 P =11/18 OPTION A
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Re: A certain dice game can only be won if, when a player throws
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