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Mary and Joe are to throw three dice each. The score is the

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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 28 Oct 2018, 04:49
hero_with_1000_faces wrote:
Bunuel

The way I did this:

values greater than 10 are 11,12,13,14,15,16,17,18. i.e 8 values
so, 8/18 (18 is the maximum Value of all three dices), so 8/18 *4/4 = 32/64.

Is this correct ?


Please read the whole thread before posting: https://gmatclub.com/forum/mary-and-joe ... ml#p788093
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Re: Mary and Joe are to throw three dice each. The score is the  [#permalink]

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New post 28 Oct 2018, 05:11
Bunuel wrote:
imania wrote:
would you love to see how attacked it?
if Joe is expected to outscore his friend, he should get these sums, 11,12,13...18
all possibilities are from 3 to 18
so : prob =8/16 equal to 1/2
PS. If you are wondering how I came to 3 as min because 1+1+1
and likewise 18 is max (6+6+6)


Unfortunately this approach is not right though for this particular case it gave a correct answer.

Consider this: if it were that Mary scored not 10 but 17 then Joe to outscore Mary should get only 18 and according to your approach as there are total of 16 scores possible then the probability of Joe getting 18 would be 1/16. But this is not correct, probability of 18 is (1/6)^3=1/216 not 1/16.

This is because not all scores from 3 to 18 have equal # of ways to occur: you can get 10 in many ways but 3 or 18 only in one way (3=1+1+1 and 18=6+6+6).

Hope it's clear.



Hi Bunuel,

Thanks for the clarification. However if the question would have asked for "number of ways Joe can get a sum greater than 13" What would have been your approach?

Please explain.
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Re: Mary and Joe are to throw three dice each. The score is the &nbs [#permalink] 28 Oct 2018, 05:11

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