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1/4
my formulas are too lengthy to be those you should actually use to solve this, so I'm waiting for other people to answer. (provided that my answer is correct!)

that is incorrect, the correct answer is 1/2...any ideas how to get to that?

damn! this is not testing our knowledge of permutations/combinations/possible outcomes
I couldn't be more stupid
Follow this:
expected value of a dice= 3.5
expected value of 3 ones= 10.5
so we have 1/2 to score less than 10.5 and 1/2 to score more than 10.5

expected value results from the sum of probability*value (it is an aleatory number)
in this case:
1/6*1+1/6*2+1/6*3+1/6*4+1/6*5+1/6*6=1/6*(1+2+3+4+5+6)=3.5
another example could be
we have 5 bottles of wine that are sold in a supermarket
2 of type A
2 of type B
1 of type C
bottles of type A are charged $5 each
bottles of type B are charged $10 each
bottles of type C are charged $15 each
John buys a bottle. If John doesn't know what bottle he chose, how much should he expect to pay?
EV=2/5*$5+2/5*$10+1/5*$15=$9

don't know if I'm clear with my English. please correct me if my reasoning is wrong

Proba is 1/2 just because we are half way from 18.
From 3 (minimum score) to 10 there is a 7 range
from 11 (minimum score to outscore) to 18 (max) you have a 7 range.

So Prob to score 11 = Prob to score 10
Prob 9 = Prob 12
...

Total possible values are from 3-18, therefore 16 outcomes.
Desired outcomes are from 11-18, therefore 8 outcomes.
Possibility is 8/16=1/2. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

There is no need for any calculations here. No matter what the numbers are, the probability will always be 1/2. The reason for that is that Joe can have only 2 outcomes- either outscore Mary or be outscored. Simple, isn't it! It would have been a different question if it was required to find out probability of Joe getting a particular score.

There is no need for any calculations here. No matter what the numbers are, the probability will always be 1/2. The reason for that is that Joe can have only 2 outcomes- either outscore Mary or be outscored. Simple, isn't it! It would have been a different question if it was required to find out probability of Joe getting a particular score.

In this case calculations are not required. because possible outcomes for sum >10 equal to possible coutcomes for sum <=10

What if question is changed to . Mary and Joe throw three dice each. If Mary gets 8, what is the probability that Joe will outscore Mary?

Definitely you need to calculate. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

min result : 3 (1+1+1) max result : 18 (6+6+6) number of outcomes : 18-3+1 = 16

We want outcome >10 and max is 18 so there are 8 possibilities over 16

8/16 = 1/2

whats the logic behind that equation?

3 dice: lowest number possible 1+1+1 =3, greatest number possible 6+6+6=18

18-3+1 (because inclusive) = 16 possible outcomes. 8 of which are greater than 10

but certain sums come up more frequently than others... ie you get 3 (1+1+1) 1/216 of the time but 4 (1+1+2,1+2+1,2+1+1) 3/216 = 1/72 of the time... _________________

There is no need for any calculations here. No matter what the numbers are, the probability will always be 1/2. The reason for that is that Joe can have only 2 outcomes- either outscore Mary or be outscored. Simple, isn't it! It would have been a different question if it was required to find out probability of Joe getting a particular score.

This is incorrect. You are right that they are two outcomes "of interest", but they are NOT either outscore Mary or be outscored. They are outscore mary and NOT be outscored (i.e., can tie). And just because there are two outcomes of interest doesn't mean they have equal probability. _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Re: Mary and Joe throw three dice each. If Mary gets 10, what [#permalink]
22 Mar 2013, 06:23

not able to understand this ques and any of the solution mentioned above.. if anyone can explain me in some other way ...then please explain thanks in advance

Re: Mary and Joe throw three dice each. If Mary gets 10, what [#permalink]
22 Mar 2013, 06:26

Expert's post

Perhaps wrote:

not able to understand this ques and any of the solution mentioned above.. if anyone can explain me in some other way ...then please explain thanks in advance

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