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Mary and Joe are to throw three dice each. The score is the [#permalink]
22 Jan 2012, 02:07

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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?

Firstly, I am aware that the following question has been discussed in other posts. But, the reason for posting this question is for me to understand where am I going wrong and also that the posts are quite old.

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. 24/64 B. 32/64 C. 36/64 D. 40/64 E. 42/64

Could somebody simply explain me the solution?

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 = 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.

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
22 Jan 2012, 04:30

2

This post received KUDOS

Expert's post

bsaikrishna wrote:

Thanks for the reply.

What if the question is to find out the probability of the sum to be greater than 12?

How can we solve it using the same expected value approach?

Since we know that the probability of getting more than 10 (11, 12, ..., 18), is 1/2 then we should find the probability of getting 11 and 12 and then subtract these values from 1/2.

But you won't need this for the GMAT as there will be lengthy calculations involved: 11 can be obtained by combination of the following: 1-4-6, 1-5-5, 2-3-6, 2-4-5, 3-4-4, 3-3-5 --> 3!+3!/2!+3!+3!+3!/2!+3!/2!=27. 12 can be obtained by combination of the following: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25.

P=1/2-(25+27)/6^3=7/27.

All combinations: The sum of 3 - 1; The sum of 4 - 3; The sum of 5 - 6; The sum of 6 - 10; The sum of 7 - 15; The sum of 8 - 21; The sum of 9 - 25; The sum of 10 - 27; The sum of 11 - 27 (notice equals to the combinations of the sum of 10); The sum of 12 - 25 (notice equals to the combinations of the sum of 9); The sum of 13 - 21 (notice equals to the combinations of the sum of 8); The sum of 14 - 15 (notice equals to the combinations of the sum of 7); The sum of 15 - 10 (notice equals to the combinations of the sum of 6); The sum of 16 - 6 (notice equals to the combinations of the sum of 5); The sum of 17 - 3 (notice equals to the combinations of the sum of 4); The sum of 18 - 1 (notice equals to the combinations of the sum of 3). Total = 2*(1+3+6+10+15+21+25+27) = 216 = 6^3.

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
05 Oct 2012, 09:30

2

This post received KUDOS

Here is a very simple conceptual techique to solve any such "sum of the dice" problems. When you throw 3 dice, what are the expected ranges of the sum of the values..?

Well (1,1,1) is the least sum and (6,6,6) is the max sum. Hence the sum can vary from 3 to 18 How many #s are in the series 3 to 18 inclusive of 18 ? Well 18-3+1 = 16 values In other words, when any one throws 3 dice, the sum of the values can either be 3,4,5,6,7,8,9....16,17,18.

For Joe to win, he has to score a sum that is more than that of Mary, ie Joe has to score between 11 and 18. How many #s are in the series 11 to 18 ? well 18-11+1 = 8

We are done Probability = Total # of favorable options / total options = 8 /16 = 32/64 = 1/2

You can use this technique for any problem that has anything to do with sum, product , division of values when multiple dies are thrown? KABOOM...

Why dont we try a different problem ? When Mary throws 3 dice, she scores 4. Joe throws a die. What is the probability that the sum of values of the dices on Joe's throw will add up to a prime number that is greater than Mary's score. _________________

----------------------------------------------------------------------------------------------------- IT TAKES QUITE A BIT OF TIME AND TO POST DETAILED RESPONSES. YOUR KUDOS IS VERY MUCH APPRECIATED -----------------------------------------------------------------------------------------------------

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
29 Sep 2012, 21:52

1

This post received KUDOS

Expert's post

stne wrote:

Bunuel I know this can be a lengthy,but can you show how the different combinations of 9 is equal to 25 I am getting 28 also the different combinations of 10 I am getting 36 well as it should be 27

the combinations from 3 to 8 matches with yours but for 9 and 10 I am getting a different answer.

Thank you

Responding to a pm:

The method you are using is not correct. It is fine for the sum till 8. It fails for 9, 10, 11 and 12. If you enumerate, you will get the same numbers as Bunuel.

First let me point out that when you decide to use a particular method, you should fully understand the method. First go through this post to understand why you can use 7C5 or 7C2 to get a sum of 8 (and to get the smaller sums too).

Notice how you divide n identical objects among m distinct groups. Let’s take the example of a sum of 7. You have to divide 7 among 3 dice such that each die must have at least 1 (no die face can show 0). First step is to take 3 out of the 7 and give one each to the three dice. Now you have 4 left to distribute among 3 distinct groups such that it is possible that some groups may get none of the four. Think of partitioning 4 in 3 groups. This can be done in (4+2)!/4!*2! = 6C2 ways (check out the given link if you do not understand this)

This is how you obtain 6C2 (which is the same as 6C4) for the sum of 7.

The concept works perfectly till the sum of 8. Thereafter it fails. Now that you know why this method works for some values, can you guess why it fails for others? _________________

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
01 Oct 2012, 21:12

1

This post received KUDOS

Expert's post

stne wrote:

maybe something to do with the minimum and maximum value of a dice? Till 8 we can have minimum 1 and maximum 6 for a dice , 1 1 6 * \frac {3!}{2} , 3 2 3 *\frac {3!}{2} , 5 2 1 * 3!,2 2 4 *\frac {3!}{2} , 1 3 4*3! = 3 +3 + 6 + 3 + 6 = 21

but for 9 we have only 1 6 2 * 3! , 1 3 5 *3!, 1 4 4 *\frac {3!}{2}, 2 2 5 *\frac {3!}{2} , 3 3 3 , 2 4 3 * 3! = 6+6+3+3+1+6 = 25

but when we use 8C6 or 8C2 for the sum of 9 , we are getting 3 extra cases, what are those extra cases ? If we can find those cases I think we can find the answer to your question.

I guess you are on the right track here. In case of 9, you give 1 to each of the 3 dice and you are left with 6. Now when you try to split 6 among the three groups, you will have 3 cases which look like this: 6, 0, 0 0, 6, 0 0, 0, 6

What you are saying here is that first die shows 7, second shows 1 and third shows 1. This case doesn't work, does it? All these 3 cases don't work and you need to remove them. _________________

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
04 Oct 2012, 02:10

1

This post received KUDOS

Expert's post

wisc4llin wrote:

Joe's possible outcomes range from 3 (if he throws all 1) to 18 (if he throws all 6).

Each option, from 3-18, is equally probable if we assume a fair dice. There are 8 outcomes (3, 4, 5, 6, 7, 8, 9, 10) where Joe gets 10 or less, thereby underscoring or tying Mary. There are 8 outcomes (11, 12... 18) in which Joe outscores Mary.

Because there are 8 ways for Joe to outscore, and 8 ways for Joe to underscore or tie and they are all equally probable, there is a 50% chance Joe will outscore Mary.

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: Mary and Joe are to throw three dice each. The score is the [#permalink]
23 Sep 2012, 07:17

Bunuel wrote:

bsaikrishna wrote:

Thanks for the reply.

What if the question is to find out the probability of the sum to be greater than 12?

How can we solve it using the same expected value approach?

Since we know that the probability of getting more than 10 (11, 12, ..., 18), is 1/2 then we should find the probability of getting 11 and 12 and then subtract these values from 1/2.

But you won't need this for the GMAT as there will be lengthy calculations involved: 11 can be obtained by combination of the following: 1-4-6, 1-5-5, 2-3-6, 2-4-5, 3-4-4, 3-3-5 --> 3!+3!/2!+3!+3!+3!/2!+3!/2!=27. 12 can be obtained by combination of the following: 1-5-6, 2-4-6, 2-5-5, 3-3-6, 3-4-5, 4-4-4 --> 3!+3!+3!/2!+3!/2!+3!+1=25.

P=1/2-(25+27)/6^3=7/27.

All combinations: The sum of 3 - 1; The sum of 4 - 3; The sum of 5 - 6; The sum of 6 - 10; The sum of 7 - 15; The sum of 8 - 21; The sum of 9 - 25; The sum of 10 - 27; The sum of 11 - 27 (notice equals to the combinations of the sum of 10); The sum of 12 - 25 (notice equals to the combinations of the sum of 9); The sum of 13 - 21 (notice equals to the combinations of the sum of 8); The sum of 14 - 15 (notice equals to the combinations of the sum of 7); The sum of 15 - 10 (notice equals to the combinations of the sum of 6); The sum of 16 - 6 (notice equals to the combinations of the sum of 5); The sum of 17 - 3 (notice equals to the combinations of the sum of 4); The sum of 18 - 1 (notice equals to the combinations of the sum of 3). Total = 2*(1+3+6+10+15+21+25+27) = 216 = 6^3.

Hope it's clear.

Bunuel I know this can be a lengthy,but can you show how the different combinations of 9 is equal to 25 I am getting 28 also the different combinations of 10 I am getting 36 well as it should be 27

the combinations from 3 to 8 matches with yours but for 9 and 10 I am getting a different answer.

Have also understood how we are able to use combinations , to arrive at the number of combinations for a particular sum of 3 dice throws ( or 2 etc).

so as explained by you, lets look at the number of combinations for the sum of 8 when 3 dices are thrown .

question was how are we able to use 7C5 to arrive at 21 combinations ?

but how come we are not able to use 8C6 to arrive at 25, 8C6 will give 28 which is three more than the number of actual combinations.

so basically till the sum of 8 , it fine but above that it seems that the method does not work.

for the sum of 8 what exactly are we doing , we have 3 dices .So lets give 1 to each , now we have to distribute 5 among 3 dices. so following the slit method we get 7C2 ( or 7C5 ) = 21 which gives the number of combinations for the sum of 8 when 3 dices are thrown.

So if the question was find the probability that the sum of 3 dices is 8 we can do = \frac{21}{6^3}.

now for the case of 9 lets try same way , give one to each dice , now we are left with 6, lets distribute 6 among 3 dices so again using the slit method we get 8C2 ( or 8C6)= 28 however we know that this is incorrect.

karishma as you can see in spite of my best efforts it seems I still cannot understand why the slit method works till the sum of 8 but not beyond that. we can distribute 6 among 3 dices can't we? 2 2 2 , 1 4 1, etc .

Well I think you have to answer this. _________________

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
01 Oct 2012, 08:44

maybe something to do with the minimum and maximum value of a dice? Till 8 we can have minimum 1 and maximum 6 for a dice , 1 1 6 * \frac {3!}{2} , 3 2 3 *\frac {3!}{2} , 5 2 1 * 3!,2 2 4 *\frac {3!}{2} , 1 3 4*3! = 3 +3 + 6 + 3 + 6 = 21

but for 9 we have only 1 6 2 * 3! , 1 3 5 *3!, 1 4 4 *\frac {3!}{2}, 2 2 5 *\frac {3!}{2} , 3 3 3 , 2 4 3 * 3! = 6+6+3+3+1+6 = 25

but when we use 8C6 or 8C2 for the sum of 9 , we are getting 3 extra cases, what are those extra cases ? If we can find those cases I think we can find the answer to your question. _________________

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
01 Oct 2012, 23:24

VeritasPrepKarishma wrote:

stne wrote:

maybe something to do with the minimum and maximum value of a dice? Till 8 we can have minimum 1 and maximum 6 for a dice , 1 1 6 * \frac {3!}{2} , 3 2 3 *\frac {3!}{2} , 5 2 1 * 3!,2 2 4 *\frac {3!}{2} , 1 3 4*3! = 3 +3 + 6 + 3 + 6 = 21

but for 9 we have only 1 6 2 * 3! , 1 3 5 *3!, 1 4 4 *\frac {3!}{2}, 2 2 5 *\frac {3!}{2} , 3 3 3 , 2 4 3 * 3! = 6+6+3+3+1+6 = 25

but when we use 8C6 or 8C2 for the sum of 9 , we are getting 3 extra cases, what are those extra cases ? If we can find those cases I think we can find the answer to your question.

I guess you are on the right track here. In case of 9, you give 1 to each of the 3 dice and you are left with 6. Now when you try to split 6 among the three groups, you will have 3 cases which look like this: 6, 0, 0 0, 6, 0 0, 0, 6

What you are saying here is that first die shows 7, second shows 1 and third shows 1. This case doesn't work, does it? All these 3 cases don't work and you need to remove them.

Thank you karishma, I am better informed now. This indeed helped. _________________

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
03 Oct 2012, 13:03

Joe's possible outcomes range from 3 (if he throws all 1) to 18 (if he throws all 6).

Each option, from 3-18, is equally probable if we assume a fair dice. There are 8 outcomes (3, 4, 5, 6, 7, 8, 9, 10) where Joe gets 10 or less, thereby underscoring or tying Mary. There are 8 outcomes (11, 12... 18) in which Joe outscores Mary.

Because there are 8 ways for Joe to outscore, and 8 ways for Joe to underscore or tie and they are all equally probable, there is a 50% chance Joe will outscore Mary.

Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
05 Oct 2012, 09:38

Expert's post

hafizkarim wrote:

Here is a very simple conceptual techique to solve any such "sum of the dice" problems. When you throw 3 dice, what are the expected ranges of the sum of the values..?

Well (1,1,1) is the least sum and (6,6,6) is the max sum. Hence the sum can vary from 3 to 18 How many #s are in the series 3 to 18 inclusive of 18 ? Well 18-3+1 = 16 values In other words, when any one throws 3 dice, the sum of the values can either be 3,4,5,6,7,8,9....16,17,18.

For Joe to win, he has to score a sum that is more than that of Mary, ie Joe has to score between 11 and 18. How many #s are in the series 11 to 18 ? well 18-11+1 = 8

We are done Probability = Total # of favorable options / total options = 8 /16 = 32/64 = 1/2

You can use this technique for any problem that has anything to do with sum, product , division of values when multiple dies are thrown? KABOOM...

Why dont we try a different problem ? When Mary throws 3 dice, she scores 4. Joe throws a die. What is the probability that the sum of values of the dices on Joe's throw will add up to a prime number that is greater than Mary's score.

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