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Re M0821 [#permalink]
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16 Sep 2014, 00:37
Official Solution:Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his? A. \(\frac{24}{64}\) B. \(\frac{32}{64}\) C. \(\frac{36}{64}\) D. \(\frac{40}{64}\) E. \(\frac{42}{64}\) To outscore Mary, Joe has to score in the range of 1118. The probability to score 3 is the same as the probability to score 18 (111 combination against 666, if 111 is on the tops of the dice the 666 is on the bottoms). By the same logic, the probability to score \(x\) is the same as the probability to score \(21  x\). Therefore, the probability to score in the range 1118 equals the probability to score in the range of 310. As 318 covers all possible outcomes the probability to score in the range 1118 is \(\frac{1}{2}\) or \(\frac{32}{64}\). Alternative Explanation Expected value of one die is \(\frac{1}{6}*(1+2+3+4+5+6)=3.5\). Expected value of three dice is \(3*3.5=10.5\). Mary scored 10 so the probability to get 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. Answer: B
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Bunuel can you please how the number 21 is derived in the original explanation? Also, here we know that the average is 10.5 & not 10 so how are we deducing that half the times sum will be less than 10 ( & not 10.5 ) & half the time ( sum will be more than 10.5) . I can understand since number showing up on the dice being integers we will never get 10.5 ( it will be either 10 or 11 but never 10.5). Nonetheless, technically speaking do we need to say that half the times sum will be greater than average & half the time it will be lesser.



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Re: M0821 [#permalink]
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08 Dec 2015, 02:37
ankushbagwale wrote: Bunuel can you please how the number 21 is derived in the original explanation? Also, here we know that the average is 10.5 & not 10 so how are we deducing that half the times sum will be less than 10 ( & not 10.5 ) & half the time ( sum will be more than 10.5) . I can understand since number showing up on the dice being integers we will never get 10.5 ( it will be either 10 or 11 but never 10.5). Nonetheless, technically speaking do we need to say that half the times sum will be greater than average & half the time it will be lesser. The probability of scoring 1 + 1 + 1 = 3 is the same as the probability of scoring 6 + 6 + 6 = 18 = 21  3. The probability of scoring 2 + 1 + 1 = 4 is the same as the probability of scoring 6 + 6 + 5 = 17 = 21  4. ...
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Re: M0821 [#permalink]
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12 Apr 2016, 05:02
Bunuel wrote: Official Solution:
Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?
A. \(\frac{24}{64}\) B. \(\frac{32}{64}\) C. \(\frac{36}{64}\) D. \(\frac{40}{64}\) E. \(\frac{42}{64}\)
To outscore Mary, Joe has to score in the range of 1118. The probability to score 3 is the same as the probability to score 18 (111 combination against 666, if 111 is on the tops of the dice the 666 is on the bottoms). By the same logic, the probability to score \(x\) is the same as the probability to score \(21  x\). Therefore, the probability to score in the range 1118 equals the probability to score in the range of 310. As 318 covers all possible outcomes the probability to score in the range 1118 is \(\frac{1}{2}\) or \(\frac{32}{64}\). Alternative Explanation Expected value of one die is \(\frac{1}{6}*(1+2+3+4+5+6)=3.5\). Expected value of three dice is \(3*3.5=10.5\). Mary scored 10 so the probability to get 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.
Answer: B Can you please list out 32 scenarios? I tried to list out but found only 30 as below: 146,156,166 236,246,256,266 326,336,346,356,366 416,426,436,446,456,466 516,526,536,546,556,566 616,626,636,646,656,666
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Re: M0821 [#permalink]
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28 May 2016, 20:39
Hi  Can someone explain a bit more about the "expected value"? What does this math mean?



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Re: M0821 [#permalink]
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28 May 2016, 21:24
glochou wrote: Hi  Can someone explain a bit more about the "expected value"? What does this math mean? Hi, when you throw a dice, it can give you any value 1 to 6.... since any of the six faces can be facing up, the probability of each was 1/6.... so total value taking this probability is\(\frac{1}{6} *1 + \frac{1}{6} *2+ \frac{1}{6} *3 + \frac{1}{6} *4+ \frac{1}{6} *5 + \frac{1}{6} *6\) = \(\frac{1}{6}(1+2+3+4+5+6)\) = \(\frac{21}{6}\) = 3.5 this is the expected value on each trow
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Re M0821 [#permalink]
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22 Jun 2016, 23:04
I think this is a highquality question and I agree with explanation. I solved like this :
Possible score values for M : 3,4,5,6,7,8,9,10 Possible score values for J : 11,12,13,14,15,16,17,18 > 8 favorable events
P(Joe will Outscore M ) = number of favorable conditions / Total number of conditions.
= 8/16 = 1/2 => 32/64
Considering my approach is right, I was wondering why this question falls under 95% hard or 700 + level question. It triggered to me like a 500 level question.



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Re: M0821 [#permalink]
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23 Jun 2016, 00:25
dharan wrote: I think this is a highquality question and I agree with explanation. I solved like this :
Possible score values for M : 3,4,5,6,7,8,9,10 Possible score values for J : 11,12,13,14,15,16,17,18 > 8 favorable events
P(Joe will Outscore M ) = number of favorable conditions / Total number of conditions.
= 8/16 = 1/2 => 32/64
Considering my approach is right, I was wondering why this question falls under 95% hard or 700 + level question. It triggered to me like a 500 level question. The logic is not correct. Check here: maryandjoearetothrowthreediceeachthescoreisthe86407.html#p788093
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Re M0821 [#permalink]
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31 Aug 2016, 05:21
I think this is a highquality question and I agree with explanation.



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Re M0821 [#permalink]
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12 Nov 2016, 05:21
I think this is a highquality question and the explanation isn't clear enough, please elaborate. Hi, Please help me understand the explanation. I did not understand the relevance of Expected Value and how does it help us in solving the question. ? Also, what if Mary scored 13 and we need to find Probability of having score more than 13, how would we solve then? Pls help. Thanks.



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Re: M0821 [#permalink]
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14 Apr 2017, 18:26
Possible score values for M : 3,4,5,6,7,8,9,10 Possible score values for J : 11,12,13,14,15,16,17,18 > 8 favorable events Is this a suitable way of attempting the Q...Probability of individual events are different right.Because it is evenly distributed this method can be applied...For a general Q can this be used??
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Re: M0821 [#permalink]
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15 Apr 2017, 02:40
VyshakhR1995 wrote: Possible score values for M : 3,4,5,6,7,8,9,10 Possible score values for J : 11,12,13,14,15,16,17,18 > 8 favorable events Is this a suitable way of attempting the Q...Probability of individual events are different right.Because it is evenly distributed this method can be applied...For a general Q can this be used?? Check here: http://gmatclub.com/forum/maryandjoe ... ml#p788093
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Re: M0821 [#permalink]
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18 Nov 2017, 08:34
please confirm if following method is correct:
to get a sum more than 10, two dices shall have values 4 or more and the third shall have 3 or more 1st dice: 4 or more probab = 3/6
2nd dice: 4 or more probab = 3/6
3rd dice: 3 ore more probab = 4/6
now final probability = \(\frac{3}{6}*\frac{3}{6}*\frac{4}{6}*\frac{3*2}{2}=\frac{1}{2}\)
multiplied by \(\frac{3*2}{2}\) as there are 3*2 ways in the 3 dice and 2 are similar so divide by 2 so final will be \(\frac{3*2}{2}\)



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Re: M0821 [#permalink]
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19 Nov 2017, 00:38
asthagupta wrote: please confirm if following method is correct:
to get a sum more than 10, two dices shall have values 4 or more and the third shall have 3 or more 1st dice: 4 or more probab = 3/6
2nd dice: 4 or more probab = 3/6
3rd dice: 3 ore more probab = 4/6
now final probability = \(\frac{3}{6}*\frac{3}{6}*\frac{4}{6}*\frac{3*2}{2}=\frac{1}{2}\)
multiplied by \(\frac{3*2}{2}\) as there are 3*2 ways in the 3 dice and 2 are similar so divide by 2 so final will be \(\frac{3*2}{2}\) Are you sure about coloured portion.. what about two as 3 and 3rd as 5 or 6.. 3,3,5... total 11
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Re M0821 [#permalink]
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10 Jul 2018, 05:08
I think this is a highquality question and I agree with explanation.










