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Mary and Joe are to throw three dice each. The score is the [#permalink]
05 Nov 2009, 13:41

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Question Stats:

25% (03:14) correct
75% (02:14) wrong based on 12 sessions

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?
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Re: mother Mary comes to me [#permalink]
05 Nov 2009, 13:59

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noboru wrote:

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?

Expected value of one die is 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 then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

Re: mother Mary comes to me [#permalink]
05 Nov 2009, 14:20

Bunuel wrote:

noboru wrote:

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?

Expected value of one die is 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 then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

Re: mother Mary comes to me [#permalink]
24 Sep 2010, 11:17

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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)

Re: mother Mary comes to me [#permalink]
24 Sep 2010, 12:28

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)

Re: mother Mary comes to me [#permalink]
24 Sep 2010, 22:20

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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).

Re: mother Mary comes to me [#permalink]
25 Sep 2010, 04:37

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).

Re: mother Mary comes to me [#permalink]
27 Sep 2010, 14:55

Yes, but alternative approaches revolve around the same idea.

I can tell you how to reduce this problem to that of a multinomial expansion if you want, but the technique is beyond the scope of GMAT. The answer presented here is the simplest possible
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Re: mother Mary comes to me [#permalink]
05 Oct 2010, 22:05

How did you get the possible scores i.e 16 and so the probablity is 1/16

Bunuel wrote:

imania wrote:

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).

Re: mother Mary comes to me [#permalink]
05 Oct 2010, 23:26

sanober1985 wrote:

How did you get the possible scores i.e 16 and so the probablity is 1/16

Bunuel wrote:

imania wrote:

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.

The possible scores are {3,4,5,...,18} which is 16 distinct numbers

But probability is NOT 1/16. The outcomes are not equally likely
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Re: mother Mary comes to me [#permalink]
06 Oct 2010, 02:05

Bunuel wrote:

noboru wrote:

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?

Expected value of one die is 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 then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

P=1/2.

Amazing explanation, but is this a GMAT type question, if yes then I doubt I will ever be able to solve such questions in Real GMAT Time and space. It is too far fetched for me to even think I can crack such a question in normal finite time, forget GMAT Time !!!
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Please give me kudos, if you like the above post. Thanks.

Re: mother Mary comes to me [#permalink]
06 Oct 2010, 02:46

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sanober1985 wrote:

How did you get the possible scores i.e 16 and so the probablity is 1/16

Bunuel wrote:

imania wrote:

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.

When you roll 3 dice you can have the following sums: 3 (min possible 1+1+1), 4, 5, 6, ...., 18 (max possible 6+6+6), so total of 16 possible sums. But as you can see in my previous post (the one you quote) the probability of these score are not equal, so it's not 1/16 for each.

devashish wrote:

Bunuel wrote:

noboru wrote:

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?

Expected value of one die is 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 then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

P=1/2.

Amazing explanation, but is this a GMAT type question, if yes then I doubt I will ever be able to solve such questions in Real GMAT Time and space. It is too far fetched for me to even think I can crack such a question in normal finite time, forget GMAT Time !!!

Don't worry, you won't see such kind of question on GMAT.
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Re: mother Mary comes to me [#permalink]
31 Jan 2011, 16:23

How were you able to come up with (1+2+3+4+5+6)? I understand that one outcome out of six occurs when Joe rolls the dice but the other part... a bit puzzling???
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

Re: mother Mary comes to me [#permalink]
31 Jan 2011, 16:36

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mariyea wrote:

How were you able to come up with (1+2+3+4+5+6)? I understand that one outcome out of six occurs when Joe rolls the dice but the other part... a bit puzzling???

Expected value of a roll of one die is 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.
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Re: mother Mary comes to me [#permalink]
01 Feb 2011, 05:45

Bunuel wrote:

mariyea wrote:

How were you able to come up with (1+2+3+4+5+6)? I understand that one outcome out of six occurs when Joe rolls the dice but the other part... a bit puzzling???

Expected value of a roll of one die is 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.

I get it now Thanks!
_________________

Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
29 May 2012, 06:42

Bunuel, was just solving this sum, what do you mean by expected value of the sum? This is a new approach for me. Would be very nice if you could explain.

I solved it using combination. to get more than 10 you need 3,3,5 and above on the dice. for ist dice you can have 3 nos more than 3, likewise for second. For third the nos are 5 &6 ie. 2 nos. so 3*3*2 = 18 no. of sums that will deliver 10+ also there are various arrangements of these 3 nos is 18*3! = 108.

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
03 Jul 2012, 02:18

vibhav wrote:

Bunuel, was just solving this sum, what do you mean by expected value of the sum? This is a new approach for me. Would be very nice if you could explain.

I solved it using combination. to get more than 10 you need 3,3,5 and above on the dice. for ist dice you can have 3 nos more than 3, likewise for second. For third the nos are 5 &6 ie. 2 nos. so 3*3*2 = 18 no. of sums that will deliver 10+ also there are various arrangements of these 3 nos is 18*3! = 108.

Total no of sums possible = 6*6*6 = 216

Probability = 108/216 = 1/2

Am i right in this approach?

This approach may not work. Assume that on first die joe got 1, second die 6 and thrid die 6, then sum is 13. hence assuming that you need 3 on first and sencond die is wrong.

at least 1 die should have 4 or more to get the sum above 10. No restriction on minimum on one die.

Bunnel's approach is right. possible outcome above 10 are 8 and possible out come below 10 are also 8. hence probability = 8/16 = 1/2
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Re: mother Mary comes to me [#permalink]
03 Jul 2012, 23:05

Bunuel wrote:

noboru wrote:

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?

Expected value of one die is 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 then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

P=1/2.

Can someone please explain what mistake i'm doing:

Re: mother Mary comes to me [#permalink]
04 Jul 2012, 00:11

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MacFauz wrote:

Bunuel wrote:

noboru wrote:

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?

Expected value of one die is 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 then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.

P=1/2.

Can someone please explain what mistake i'm doing:

So, total of 60 scenarios were missing. Together with the 48 cases you counted we would have 48+60=108 ways to get the sum of 10 or less, so the probability is 1-108/216=1/2.