Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: mother Mary comes to me [#permalink]
01 Oct 2012, 02:52
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.
Hi Bunel, I used below appraoch to get to thr solution. Let me know if this is correct.
This problem can be drilled down to find all possible ways to get sum 11 and above.
Dice 1 : 6 , Dice2 : 4 and Dice3 : {1,2,3,4,5,6} This gives total 6 combinations(1 x 1 x 6). Now these values can be interchanged on other dices so = 6 x 3 x 2 x 1. So Combination 1 = 36 ways
Step2 : Similarly lets try to get values 6,5 and {1,2,3,4,5,6} Therefore Combination 2 = 36 ways
Step3 : Similarly lets try to get values 6,6 and {1,2,3,4,5,6} Therefore Combination 3 = 36 ways
Re: mother Mary comes to me [#permalink]
05 Nov 2012, 18:53
navo 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?
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.
Hi Bunel, I used below appraoch to get to thr solution. Let me know if this is correct.
This problem can be drilled down to find all possible ways to get sum 11 and above.
Dice 1 : 6 , Dice2 : 4 and Dice3 : {1,2,3,4,5,6} This gives total 6 combinations(1 x 1 x 6). Now these values can be interchanged on other dices so = 6 x 3 x 2 x 1. So Combination 1 = 36 ways
Step2 : Similarly lets try to get values 6,5 and {1,2,3,4,5,6} Therefore Combination 2 = 36 ways
Step3 : Similarly lets try to get values 6,6 and {1,2,3,4,5,6} Therefore Combination 3 = 36 ways
Therefore in all we have 36 + 36 + 36 = 108 ways.
and total number of ways = 216
Therefore probablity = 108/216.
It is not clear what you mean by "Now these values can be interchanged on other dices so = 6 x 3 x 2 x 1." What about combinations {5,5,5}, {4,4,4} for example? I don't see they are counted in your approach.
I also tried to solve the task by counting combinations which would give 10 points or less with the plan to substract this number from total number of combinations. Then I realized that this value would be the same as if I would count number of combinations resulting in 11 points and higher. Actually we dont even need to find average expected value.
There are 16 possible scores: 3,4,5,...16,17,18. Work from opposite sides of this set. Possibility to get 3 points is the same as for 18, possibility of 4 is the same as of 17, and so on. The last pair in the middle is 10 and 11. So possibility of score from 3 to 10 = possibility 11 to 18.
The question is about getting score more than 11, i.e. second half of the set = exactly half of choices.
similarly for 14,13,12,and 11 we have 15,21,25,27 possibilities respectively. Total favorable: 1+3+6+10+15+21+25+27 = 108 possibilities
Probability = 108/(6*6*6) = 1/2 or 32/64
Two important points... the solution is not the shortest but shows systematic listing method useful for other questions. Secondly, It appeared as though a series was forming.. which is not the case!!!
Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
05 Jan 2013, 06:57
Short-cut solution:
3 is minimum and 18 is maximum as the sum of three die. Now, 18 will have same possibilities as 3 (1 possibility) 17 will have same possibilities as 4 (3 possibilites) 16 will have same possibilities as 5 (6 possibilites) . . . 10 will have same possibilities as 11 (27 possibilites)
NOTE: 18+3 = 21, 17 + 4 = 21.... this helps to generate the series.
Thus all possibilites are half - half divided between 3-10 and 11-18. Thus if Joe has to get something from 11 to 18, his probability is 1/2
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.
Hope it helps.
My brain works in terms of combinatorics better than the "fast" approach. I feel it is much more intuitive. It is tough to come with the fast one immediately - it is good (My hats off to Bunuel and others who explain, get, and use this approach) but in my view it is more of a "you either come up with it or nothing" and therefore, completely unreliable (and just remembering solution to this problem will not get you the right answer once the similar problem is tweaked in a slightly different manner, as is the case with most GMAT problems). Combinations is a more conceptual and a step by step way that will and should eventually get you there. Now, I was wondering can a combinations way be summarized for this problem? Or what can be the most concise comb way to approach this? Ironically, to me this is one of the instances when shorter does not necessarily mean better, not to offend anyone here. _________________
There are times when I do not mind kudos...I do enjoy giving some for help
Re: mother Mary comes to me [#permalink]
29 Mar 2014, 21:53
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.
@Bunuel Plz help me with this. My method to solve was as below. Where did I go wrong?
Joe should score more than 10 .Possible cases {4,3,4}* 3!/2!(ways of arrangement) , {3,4,5}*3! , {4,4,4}*1 ,{5,5,5}*1 ,{6,6,6}*1 Probability of getting any of the above case is (1/6)^3 .. Thus, [(1/6)^3 * 3!] + [(1/6)^3 * 3!/2!] + [(1/6)^3 * 1] + [(1/6)^3 * 1] +[(1/6)^3 * 1] = 1/18 ..
Re: mother Mary comes to me [#permalink]
30 Mar 2014, 10:39
Expert's post
sachinwar 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.
@Bunuel Plz help me with this. My method to solve was as below. Where did I go wrong?
Joe should score more than 10 .Possible cases {4,3,4}* 3!/2!(ways of arrangement) , {3,4,5}*3! , {4,4,4}*1 ,{5,5,5}*1 ,{6,6,6}*1 Probability of getting any of the above case is (1/6)^3 .. Thus, [(1/6)^3 * 3!] + [(1/6)^3 * 3!/2!] + [(1/6)^3 * 1] + [(1/6)^3 * 1] +[(1/6)^3 * 1] = 1/18 ..
Plz help !!!
You are missing several other cases: {6, 6, 5} {6, 6, 4} {6, 6, 3} {6, 6, 2} ... _________________
Re: mother Mary comes to me [#permalink]
27 Apr 2014, 10:46
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.
Sorry about this ..I m not too good in Probability.
How did you come up with multiplied by 1 , 2 and so on ..
To Roll a die .. possible outcome for each will be 1/6 .. but you multiplied this by 1 then 2 then 3 and so on .. means 1/6*1 .what is multiplied by 1 denotes here ?
Re: mother Mary comes to me [#permalink]
28 Apr 2014, 02:48
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:
Re: mother Mary comes to me [#permalink]
28 Apr 2014, 02:50
Expert's post
himanshujovi wrote:
MacFauz wrote:
Bunuel 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: Mary and Joe are to throw three dice each. The score is the [#permalink]
30 May 2014, 00:31
1
This post received KUDOS
Expert's post
mrvora wrote:
Hi Bunuel, Can you please explain your answer of (1/6)^3 = 1/216 in case mary scores 17.
If Mary scores 17, then for Joe to outscore her should get the score of 18 (max possible with three dice: 6+6+6=18). The probability of getting 18, so the probability of getting 6 on each of the three dice, is 1/6*1/6*1/6 = (1/6)^3 = 1/216.
Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
08 Jun 2014, 22:38
If the problem were to be changed slightly and said that none of the first three digits repeat themselves, then the answer would become 3x2x2 = 12 going the the original method posted by Bunuel. Is that correct?
Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
22 Jul 2014, 14:37
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.
I think it is the only way to solve the problem in less than 2 minutes. When I tried to calculate the total probability of 11, 12, ...18, it took me more than 10 minutes. _________________
......................................................................... +1 Kudos please, if you like my post
Mary and Joe are to throw three dice each. The score is the [#permalink]
20 Aug 2014, 02:01
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.
======== Thanks Bunuel for the explanation however my 2 pence are as follows , The original explanation is perfectly sound, kindly correct me if I am wrong.
P of Joe getting more than 10 is 8 ways / 16 ways =1/2 (yay!) 8= 11, 12,....18; 16= 3,4....18;
Range of Summation of scores is 16 ; Max =18 and Min = 3 [ 3=1+1+1; 18=6+6+6 ] Prob of each =1/16. lets consider an Outcome of 15 [ 3+6+6, 4+5+6............] ,
Question 1, why should be the permutations of numbers within the group be considered as the outcome is always 15 , per my example and as far as the verbiage of the question goes.... so aint these are identical groups of three since the outcome is 15 always. So each of the Outcomes i.e 16 of them [per question] shud have an equal probability i.e 1 /16.
Question 2, why are we considering the expected value when the question asks of Probability of Joe getting certain digits as outcomes. Regards..
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...