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Ram and Shyam planned to be play a game in which they
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07 Dec 2019, 19:08
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39% (02:00) correct 61% (02:44) wrong based on 31 sessions
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GMATBusters’ Quant Quiz Question 5 Ram and Shyam planned to be play a game in which they are to throw three dice each. The gamepoints is the sum of points on all three dice. If Ram scores 10 in his attempt what is the probability that Shaam will outscore Ram in his? A. 24/64 B. 32/64 C. 36/64 D. 40/64 E. 42/64
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Re: Ram and Shyam planned to be play a game in which they
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07 Dec 2019, 20:37
Answer:B
Total no of outcomes=6*6*6=216 favorable outcomes >10 ae below. Sum of 18: 6+6+6 (1 outcome meeting the criteria of sum > 10) Sum of 17: 6+6+5 (3 outcomes: 665, 656, 566) Sum of 16: 6+6+4 (3 outcomes), 6+5+5 (3 outcomes): total of 6 outcomes Sum of 15: 6+6+3 (3 outcomes), 6+5+4 (6 outcomes: 654, 645, 564, 546, 465, 456), 5+5+5 (1 outcome): total of 10 outcomes Sum of 14: 6+6+2 (3), 6+5+3 (6), 6+4+4 (3), 5+5+4 (3): total of 15 outcomes Sum of 13: 6+6+1 (3), 6+5+2 (6), 6+4+3 (6), 5+5+3 (3), 5+4+4 (3): total of 21 outcomes Sum of 12: 6+5+1 (6), 6+4+2 (6), 6+3+3 (3), 5+5+2 (3), 5+4+3 (6), 4+4+4 (1): total of 25 outcomes Sum of 11: 6+4+1 (6), 6+3+2 (6), 5+5+1 (3), 5+4+2 (6), 5+3+3 (3), 4+4+3 (3): total of 27 outcomes
Outcomes >10 = 108
108/216=1/2
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Re: Ram and Shyam planned to be play a game in which they
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08 Dec 2019, 01:15
If we see the minimum value of sum on all 3 dice its 3 and maximum value of sum on all 3 dice its 6
Since all dice has equal probability to occur. So result is directly 1/2 since probability of sum of value from 3 to 10 will be same as probability of value from 11 to 18.
Ans  B



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Re: Ram and Shyam planned to be play a game in which they
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Updated on: 09 Dec 2019, 10:34
Ans:B All the possible Scores: 3,4,5...18(Total Number of scores = 183+1=16) The median score that the dice can produce is (3 + 18)/2 = 21/2 = 10.5 In order to Win the Game Shyam has to score any of the following score(11,12,...18) There is a 50% chance of scoring (8 out of 16 possible numbers will make shyam win the game) Hence the Probablity of scoring any sum among (11,12,...18) will be 8/16 or 0.5
Originally posted by numb007 on 08 Dec 2019, 01:23.
Last edited by numb007 on 09 Dec 2019, 10:34, edited 1 time in total.



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Re: Ram and Shyam planned to be play a game in which they
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08 Dec 2019, 01:35
total fair chances ; 6^3 ; 216 and for Shyam to outscore Ram he has to score>10 ; for eg (1,5,6) ( 1,6,6)... so on total possible cases ; 84 P = 84/216 ; .38 IMO A ; 24/64
Ram and Shyam planned to be play a game in which they are to throw three dice each. The gamepoints is the sum of points on all three dice. If Ram scores 10 in his attempt what is the probability that Shyam will outscore Ram in his?
A. 24/64 B. 32/64 C. 36/64 D. 40/64 E. 42/64



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Re: Ram and Shyam planned to be play a game in which they
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Updated on: 08 Dec 2019, 22:23
Ram and Shyam planned to be play a game in which they are to throw three dice each. The gamepoints is the sum of points on all three dice. If Ram scores 10 in his attempt what is the probability that Shyam will outscore Ram in his?
A. 24/64 B. 32/64 C. 36/64 D. 40/64 E. 42/64
Expected value of one dice is 16∗(1+2+3+4+5+6)=3.5. Expected value of three dice is 3∗3.5=10.5 Ram scored 10 so the probability of shyam getting the sum more than 10 (11, 12, 13, ..., 18), or more than the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2 = 32/64. That's because the probability distribution is symmetrical for this case: The probability of getting the sum of 3 (min possible sum) = the probability of getting the sum of 18 (max possible sum); The probability of getting the sum of 4 = the probability of getting the sum of 17; The probability of getting the sum of 5 = the probability of getting the sum of 16; ........................................... The probability of getting the sum of 10 = the probability of getting the sum of 11;
Thus the probability of getting the sum from 3 to 10 = the probability of getting the sum from 11 to 18 = 1/2.
Imo. B
Originally posted by Raxit85 on 08 Dec 2019, 04:24.
Last edited by Raxit85 on 08 Dec 2019, 22:23, edited 1 time in total.



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Re: Ram and Shyam planned to be play a game in which they
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Updated on: 08 Dec 2019, 22:23
The answer is B: 32/64.
EXPLANATION: Let's first find the total number of cases. That would come to be as \( 6^3 \) = 216. This is because three dices are thrown, and each dice has 6 outcomes.
Now, we find the number of valid cases. We have to find the total cases in which Shaam can outscore Ram's 10 points. We can either find cases of >10 (which would be 11, 12, 13, 14, 15, 16, 17, 18), or cases of <=10 (3, 4, 5, 6, 7, 8, 9, 10), and subtract 1 from the probability calculated (as we would calculate cases of not outscoring Ram). Let's do the later, as it is easier.
We form the cases, for scoring below 10, upto maximum 9. (NOTE: We use the formula: n!/a!b!.., for all common values grouped).
Case 1: 3 points (minimum value) Can only be achieved by 1 way: 1, 1, 1. TOTAL: 1
Case 2: 4 points all arrangement of 1, 1, 2 = 3 TOTAL: 3
Case 3: 5 points all arrangement of 1, 2, 2 = 3 all arrangement of 1, 1, 3 = 3 TOTAL: 6
Case 4: 6 points all arrangement of 1, 1, 4 = 3 all arrangement of 1, 2, 3 = 6 all arrangement of 2, 2, 2 = 1 TOTAL: 10
Case 5: 7 points all arrangement of 1, 1, 5 = 3 all arrangement of 1, 2, 4 = 6 all arrangement of 1, 3, 3 = 3 all arrangement of 2, 2, 3 = 3 TOTAL: 15
Case 6: 8 points all arrangement of 1, 1, 6 = 3 all arrangement of 1, 2, 5 = 6 all arrangement of 1, 3, 4 = 6 all arrangement of 2, 2, 4 = 3 all arrangement of 3, 3, 2 = 3 TOTAL: 21
Case 8: 9 points all arrangement of 3, 3, 3 = 1 all arrangement of 2, 2, 5 = 3 all arrangement of 4, 4, 1 = 3 all arrangement of 1, 2, 6 = 6 all arrangement of 2, 3, 4 = 6 all arrangement of 5, 1, 3 = 6 TOTAL: 25
Case 9: 10 points all arrangement of 1, 3, 6 = 6 all arrangement of 1, 4, 5 = 6 all arrangement of 2, 2, 6 = 3 all arrangement of 2, 3, 5 = 6 all arrangement of 2, 4, 4 = 3 all arrangement of 3, 3, 4 = 3 TOTAL: 27
GRAND TOTAL OF POSSIBLE CASES: 27 + 25 + 21 + 15 + 10 + 6 + 3 + 1 = 108.
So probability = possible cases/total number of cases = 108/216, which is exactly 0.5
Note, this is the probability of NOT outscoring Ram. To outscore Ram, we subtract this with 1. So, probability of outscoring Ram = 1  0.5 = 0.5.
Among the answer choices, only B, 32/64 = 0.5 matches our answer.
Hence, answer is B: 32/64.
Originally posted by gravito123 on 08 Dec 2019, 05:58.
Last edited by gravito123 on 08 Dec 2019, 22:23, edited 1 time in total.



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Re: Ram and Shyam planned to be play a game in which they
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08 Dec 2019, 06:30
The following image shows the distribution of combinations of 3 dice in which all the possible arrangements of 3 dice are plotted against their outcome. It can be seen that to get a outcome of 3, 1 arrangement is possible. Similarly for outcome of 4, 3 arrangements are possible. For outcome of 5  6 arrangements For outcome of 6  10 arrangements For outcome of 7  15 arrangements For outcome of 8  21 arrangements For outcome of 9  25 arrangements For outcome of 10  27 arrangements For outcome of 11  27 arrangements For outcome of 12  25 arrangements For outcome of 13  21 arrangements For outcome of 14  15 arrangements For outcome of 15  10 arrangements For outcome of 16  6 arrangements For outcome of 17  3 arrangements For outcome of 18  1 arrangements Now, for Shyam to outscore Ram, Shyam must score more than 10, i.e. he can score from 11 to 18. Method I Shyam can score in the following manner. Score of 18 – 1 case P = 1/216 Score of 17 – 3 cases P = 3/216 Score of 16 – 6 cases P = 6/216 Score of 15 – 10 cases P = 10/216 Score of 14 – 15 cases P = 15/216 Score of 13 – 21 cases P = 21/216 Score of 12 – 25 cases P = 25/216 Score of 11 – 27 cases P = 27/216 Total Probabilty = 1/216 + 3/216 + 6/216 + 10/216 + 15/216 + 21/216 + 25/216 + 27/216 = 108/216 = 1/2 = 32/64 Method  II By symmetry, the sum of the number of arrangements from 3 to 10 is equal to the sum of the number of arrangements from 11 to 18. Therefore, probability, P = 1/2 = 32/64
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Re: Ram and Shyam planned to be play a game in which they
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09 Dec 2019, 05:45
Total sum for 3 dice lies from 3 to 18 For Shyam to outscore Ram, Shyam should score 11, 12, 13, 14, 15, 16, 17 or 18 Since the probability distribution is symmetric about 10 and 11 and Shyam is to score more than 10, so P = 1/2 = 32/64
B is the answer



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Re: Ram and Shyam planned to be play a game in which they
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08 Jan 2020, 04:09
gmatbusters wrote: GMATBusters’ Quant Quiz Question 5 Ram and Shyam planned to be play a game in which they are to throw three dice each. The gamepoints is the sum of points on all three dice. If Ram scores 10 in his attempt what is the probability that Shaam will outscore Ram in his? A. 24/64 B. 32/64 C. 36/64 D. 40/64 E. 42/64 My one line approach To outscore RAM, Shaam needs 11 ( 1,4,6) first dice he has 6 options, second 3 options, third 1 options and outcome on each dice can be interchanged in !3 ways Total=6*3*1*!3=108 Required Probability=108/216 =1/2 B:)




Re: Ram and Shyam planned to be play a game in which they
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