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While playing a certain dice game, Chris wins if the sum of
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10 Aug 2014, 03:58
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58% (02:28) correct 42% (02:36) wrong based on 231 sessions
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While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win? A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216 I could not understand how to approach this problem. Can someone throw some light on how to solve this.
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While playing a certain dice game, Chris wins if the sum of
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12 Aug 2014, 06:56
Total outcomes possible: 36 Total outcomes possible with sum 7: 6 Probability to win when rolled once = 6/36 Probability not to win when rolled once = 30/36 = 5/6 Probability to win in three attempts= 1 Probability will not to win in all three attempts = 1 (5/6* 5/6*5/6) = 91/216




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Re: While playing a certain dice game, Chris wins if the sum of
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10 Aug 2014, 07:46
There are 36 ways that a pair of dice can land. To have a sum of 7, there are 6 ways: 16, 25, 34, 43, 52 and 61
Therefore, the chance to have a sum of 7 is \(\frac{1}{6}\).
Making a sum of 7 on the first time is \(\frac{1}{6}=\frac{36}{216}\) Making a sum of 7 on the second time is \(\frac{5}{6}*\frac{1}{6}=\frac{5}{36}=\frac{30}{216}\) Making a sum of 7 on the third time is \(\frac{5}{6}*\frac{5}{6}*\frac{1}{6}=\frac{25}{216}\)
Add them all together to get \(\frac{91}{216}\)
Answer is C.



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Re: While playing a certain dice game, Chris wins if the sum of
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16 Aug 2014, 14:54
akhil911 wrote: While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?
A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216
I could not understand how to approach this problem. Can someone throw some light on how to solve this. I understand the solution. But I have a meaning problem: How many dices could be rolled (max)? Three pairs (6) or three dices (3) I understand that he rolls the First dice (from 1 to 6). No thing I can do. Second dice (1 to 6) = 1/6 of chance of getting a 7. If the sum is below 7: Third dice (1 to 6). I dont know how to calculate! =) My final question is: is this problem ambiguous?



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Re: While playing a certain dice game, Chris wins if the sum of
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17 Aug 2014, 22:31
plaverbach wrote: akhil911 wrote: While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?
A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216
I could not understand how to approach this problem. Can someone throw some light on how to solve this. I understand the solution. But I have a meaning problem: How many dices could be rolled (max)? Three pairs (6) or three dices (3) I understand that he rolls the First dice (from 1 to 6). No thing I can do. Second dice (1 to 6) = 1/6 of chance of getting a 7. If the sum is below 7: Third dice (1 to 6). I dont know how to calculate! =) My final question is: is this problem ambiguous? I think question sould be read as follows: Charlie has three rolls with two dices each of them. Hence every roll minimum score would be 2 (1+1) and maximum 12 (6+6). Probability of having 7 in the first roll is 1/6 (6 out of 36). You can follow one of the two ways explained here above. Both works. Hope it helps.



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While playing a certain dice game, Chris wins if the sum of
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01 Sep 2014, 00:59
Why dont be consider a 4th option where on none of the 3 rolls he gets a 7? shouldnt we be adding [5][/6]^3 to the 91/216?



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While playing a certain dice game, Chris wins if the sum of
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26 Oct 2015, 23:35
wunsun wrote: There are 36 ways that a pair of dice can land. To have a sum of 7, there are 6 ways: 16, 25, 34, 43, 52 and 61
Therefore, the chance to have a sum of 7 is \(\frac{1}{6}\).
Making a sum of 7 on the first time is \(\frac{1}{6}=\frac{36}{216}\) Making a sum of 7 on the second time is \(\frac{5}{6}*\frac{1}{6}=\frac{5}{36}=\frac{30}{216}\) Making a sum of 7 on the third time is \(\frac{5}{6}*\frac{5}{6}*\frac{1}{6}=\frac{25}{216}\)
Add them all together to get \(\frac{91}{216}\)
Answer is C. I am not able to understand above approach. (please elaborate it) however I have got answer using inclusion exclusion principle..... Considering 3 rolls of 2 dice A= event that I get sum equal to 7 in first roll B= event that I get sum equal to 7 in 2nd roll C= event that I get sum equal to 7 in 3rd roll in every roll we have six ways to get sum equal to 7 (i.e. 1,6 or 6,1 and so on) A∪B∪C = A + B + C  A∩B  B∩C  A∩C + A∩B∩C = \(1/6+1/6+1/61/6*1/61/6*1/61/6*1/6+1/6*1/6*1/6\) = \(91/216\) IS this way correct?



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Re: While playing a certain dice game, Chris wins if the sum of
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27 Oct 2015, 07:01
akhil911 wrote: While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?
A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216
I could not understand how to approach this problem. Can someone throw some light on how to solve this. Probability of winning game = 1 Probability of losing game Probability of losing game = (Probability of not getting sum 7 in any of the three attempts) Ways of getting sum 7 = (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)= 6 ways Total ways of getting outcome on two dice =6*6=36 Probability of getting sum as 7 in any attempt =6/36=1/6 Probability of NOT getting sum as 7 in any attempt = 1(1/6)= 5/6 Probability of losing game =(5/6)*(5/6)*(5/6)=125/216 I.e. Probability of wining game = 1(125/216) = 91/216 Answer : Option C
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Re: While playing a certain dice game, Chris wins if the sum of
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11 Mar 2016, 20:09
oh man..I understood that he has to throw 3 dice..not 3 times.... so i took.. probability that it won't work.. 1*5/6 x 5/6 = 25/36 and thus, 11/36 is probability that out of 3 dice, he will get 7..
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Re: While playing a certain dice game, Chris wins if the sum of
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13 Mar 2016, 08:48
akhil911 wrote: While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?
A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216
I could not understand how to approach this problem. Can someone throw some light on how to solve this. There are 6 possibilities for the sum to be 7. (1,6) (2,5) (3.4) and the three reverse order pairs. The probability of a win is 6 over 36 or 1/6 and the probability of a loss is 5/6 Now, there are three possible cases: W or LW or LLW so, 1/6 + 5/6 * 1/6 + 5/6 * 5/6 * 1/6 = 91/216



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Re: While playing a certain dice game, Chris wins if the sum of
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22 Aug 2018, 13:00
akhil911 wrote: While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?
A. 1/2 B. 17/36 C. 91/216 D. 11/36 E. 25/216
I could not understand how to approach this problem. Can someone throw some light on how to solve this. The sum of two dice can be 7 in six ways  (1,6) , (2,5), (3,4), (4,3) , (5,2) and (6,1) Total combinations of the numbers on throwing two dice = 36 Thus probability of getting a sum of 3 = 6/36 = 1/6 Probability of not getting a sum of 7 = 11/6 = 5/6 Now Chris has 3 attempts to get the desired sum. If he gets sum 7 in 1st attempt than the result is achieved and the other two attempts are not required > P1= 1/6 If Chris does not gets sum 7 in 1st attempt, then he needs another attempt and if he gets six then prob > P2= 5/6*1/6 (P(not getting sum 7 in 1st attempt)* P(get sum 7 in 2nd attempt)) Similarly, a 3rd attempt would be needed if Chris fails to gets sum 7 in previous attempts > P3 = 5/6*5/6*1/6 Total prob = P1 +P2+ P3 = 1/6 + 5/6*1/6 * 5/6*5/6*1/6 = (36+30+25)*6^3 = 91/126




Re: While playing a certain dice game, Chris wins if the sum of
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