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 Author: galiya [ 25 Apr 2012, 09:59 ] Post subject: A certain team has 12 members, including Joey. A three-member relay te A certain team has 12 members, including Joey. A three-member relay team will be selected as follows: one of the 12 members is to be chosen at random to run first, one of the remaining 11 members is to be chosen at random to run second, and one of the remaining 10 members is to be chosen at random to run third. What is the probability that Joey will be chosen to run second or third?A. 1/1,320B. 1/132C. 1/110D. 1/12E. 1/6Guysactually i have an official explanation of the right answer, but its a bit illogical for meExplanation: There is a lot of excess wording to this question when it is really a simple concept. Each of the team members has an equal chance to be selected to run first, second, or third, and (perhaps obviously) no team member can be selected to run more than one of those. Therefore,from Joey's perspective, he has a 1/12 chance of running first, a 1/12 chance of running second,and a 1/12 chance of running third. Since he can't run both second AND third,the chances that he'll run second OR third is the sum of those two probabilities: 1/12 +1/12 =2/12 =1/6but how is it possible to have a probability of 1/12, that Joey will run the second or the third, if the first runner has already started? After this we have just 11 members, and the probability should be 1/11 that Joey will start the second, accordingly 1/10th that he will run the thirdCould u please share with your thoughts on this?

 Author: Bunuel [ 25 Apr 2012, 10:37 ] Post subject: A certain team has 12 members, including Joey. A three-member relay te Galiya wrote:A certain team has 12 members, including Joey. A three-member relay team will be selected as follows: one of the 12 members is to be chosen at random to run Örst, one of the remaining 11 members is to be chosen at random to run second, and one of the remaining 10 members is to be chosen at random to run third. What is the probability that Joey will be chosen to run second or third?A. 1/1,320B. 1/132C. 1/110D. 1/12E. 1/6Guysactually i have an official explanation of the right answer, but its a bit illogical for meExplanation: There is a lot of excess wording to this question when it is really a simple concept. Each of the team members has an equal chance to be selected to run first, second, or third, and (perhaps obviously) no team member can be selected to run more than one of those. Therefore,from Joey's perspective, he has a 1/12 chance of running first, a 1/12 chance of running second,and a 1/12 chance of running third. Since he can't run both second AND third,the chances that he'll run second OR third is the sum of those two probabilities: 1/12 +1/12 =2/12 =1/6but how is it possible to have a probability of 1/12, that Joey will run the second or the third, if the first runner has already started? After this we have just 11 members, and the probability should be 1/11 that Joey will start the second, accordingly 1/10th that he will run the thirdCould u please share with your thoughts on this?Standard approach:(any but Joey)(Joey)(any) + (any but Joey)(any but Joey)(Joey) = 11/12*1/11*1+11/12*10/11*1/10=2/12.Answer: E.Another approach:Actually even OE has one more step than necessary: since there are two slots for Joey from 12 possible than the probability is simply 2/12. Consider this: line 12 members in a row. Now, what is the probability that Joey is 1st in that row? 1/12. What is the probability that he's 2nd? Again 1/12. What is the probability that he's 12th? What is the probability that he's second or third? 1/12+1/12=2/12. What is the probability that he's in last 6? 6/12...Answer: E.Hope it's clear.

 Author: Zarrolou [ 07 Jun 2013, 14:07 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te Let s say that we have three spots to fill _ _ _ one for each positionCase Joey second $$11*1*10$$, we can take 11 people for the first one, only one (Joey) for the second one and 10 of the remaining for the third place.Case Joey third $$11*10*1$$, with the same logic.The total cases possible are $$12*11*10$$, this time we consider all 12 people at the beginning, with no limitations.Probability = $$\frac{2*10*11}{10*11*12}=\frac{1}{6}$$

 Author: galiya [ 25 Apr 2012, 10:45 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te if rephrase the question "What is the probability that Joey will run second or third?", will we get the probability 1/11+1/10?

 Author: Bunuel [ 25 Apr 2012, 10:54 ] Post subject: A certain team has 12 members, including Joey. A three-member relay te Galiya wrote:if rephrase the question "What is the probability that Joey will run second or third?", will we get the probability 1/11+1/10?No, it the same question. If you are doing this way, you should account there that to be second he shouldn't run first (11/12) and to be third he shouldn't run first or second (11/12*10/11): P= 11/12*1/11*1+11/12*10/11*1/10=2/12 (standard approach from above).Check similar questions to practice:https://gmatclub.com/forum/a-certain-tel ... 27423.htmlhttps://gmatclub.com/forum/a-medical-res ... 27396.htmlhttps://gmatclub.com/forum/a-certain-cla ... 27730.htmlHope it helps.

 Author: galiya [ 26 Apr 2012, 12:05 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te Bunueli went thru the add stuff, you had provided.FTB I'm absolutely confused by this kind of questions!Could you please give me an example where it will be appropriate to use the method with decreasing denominators (1/11+1/10) in contrast to the one above?eg the probability wouldn't be the same for the each member within the team.I need to clarify for myself when to use what approach

 Author: Bunuel [ 26 Apr 2012, 12:54 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te Galiya wrote:Bunueli went thru the add stuff, you had provided.FTB I'm absolutely confused by this kind of questions!Could you please give me an example where it will be appropriate to use the method with decreasing denominators (1/11+1/10) in contrast to the one above?eg the probability wouldn't be the same for the each member within the team.I need to clarify for myself when to use what approachGo through the questions here: search.php?search_id=tag&tag_id=54There should be a lot of the questions of that type.

 Author: Bunuel [ 07 Jun 2013, 06:03 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te Bumping for review and further discussion*. Get a kudos point for an alternative solution!*New project from GMAT Club!!! Check HERETheory on probability problems: math-probability-87244.htmlAll DS probability problems to practice: search.php?search_id=tag&tag_id=33All PS probability problems to practice: search.php?search_id=tag&tag_id=54Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

 Author: lchen [ 07 Jun 2013, 07:15 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te Bunuel, don't you mean answer is E?Here is another way to solve it.1 - (chance Joey will run first) - (chance Joey will not run at all)$$1 - \frac{1}{12} - (\frac{11}{12} * \frac{10}{11} * \frac{9}{10})$$$$1 - \frac{1}{12} - \frac{9}{12}$$$$1 - \frac{10}{12}$$$$\frac{2}{12} = \frac{1}{6}$$Answer is EExplanation: If Joey doesn't run first and he actually gets the chance to run, that means that he has to run either second or third.

 Author: vivmechster [ 31 Oct 2013, 21:32 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te probability that Joey will be chosen to run second or thirdMeans, Chosen Second = Not chosen first * chosen second -------------------(A)Chosen third = Not chosen first * not chosen second * chosen third. --------(B)Total: (A)+(B) = (11/12)*(1/11) + (11/12)*(10/11)*(1/10)=1/6 Hence E

 Author: koreye [ 26 Sep 2015, 10:25 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te For an easy way to look at it and not be confused with how the probability can be 1/12 in each instance, think of it this way:The question is asking the probability of either one or the other happening. You have to find the probability that Joey will either be chosen second to run or the third.Case 1 (Prob of Joey is chosen to run second)= Probability that anyone except Joey will be chosen to run first (11/12) * Probability that Joey will be chosen to Run second (1/11) * Probability that any one of the rest will be chosen to run third (10/10)Which is = 11/12 * 1/11 * 10/10 = 1/12Case 2 (Probability that Joey will be chosen to run third) = Probability that anyone except Joey will be chosen to run first (11/12) * Probability that anyone except Joey will be chosen to run second (10/11) * Probability that Joey will be chosen to run third (1/10)= 11/12 * 10/11 * 1/10 = 1/12Now to find the answer, add the two probabilities (as we do in either or cases) to find the probability that Joey will either be chosen second to run or the third = 1/12 + 1/12 = 1/6Answer: E

 Author: ScottTargetTestPrep [ 18 Sep 2019, 09:01 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te galiya wrote:A certain team has 12 members, including Joey. A three-member relay team will be selected as follows: one of the 12 members is to be chosen at random to run first, one of the remaining 11 members is to be chosen at random to run second, and one of the remaining 10 members is to be chosen at random to run third. What is the probability that Joey will be chosen to run second or third?A. 1/1,320B. 1/132C. 1/110D. 1/12E. 1/6[The probability Joey will be chosen to run second is:11/12 x 1/11 x 10/10 = 1/12The probability Joey will be chosen to run third is:11/12 x 10/11 x 1/10 = 1/12Thus, the probability that he will be chosen to run second or third is:1/12 + 1/12 = 2/12 = 1/6Answer: E

 Author: kornn [ 13 Jun 2020, 18:55 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te Dear IanStewart VeritasKarishma GMATGuruNY,Q1. If this question were replacement problem, would the answer be 1/12 + 1/12 - (1/12)*(1/12) ?Q2. Since the original question is NON-replacement (it's like choosing 3 positions SIMULTANEOUSLY), it is mutually exclusive (meaning that Joey can't run second and third at the same time). Hence, the (1/12)*(1/12) part disappears?Q3. We can add together 1/12 + 1/12 ONLY when all the 12 members are DISTINCT, right? This approach cannot be generalized to, say, 4 red marbles and 8 blue marbles where there are repeats.

 Author: IanStewart [ 14 Jun 2020, 03:22 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te varotkorn wrote:Q1. If this question were replacement problem, would the answer be 1/12 + 1/12 - (1/12)*(1/12) ?Yes, you're using Venn diagram principles, and you can solve that way. But you'll have a lot more flexibility (i.e. you'll be able to solve a wider variety of problems) if you can think about this using probability principles. If we're selecting with replacement (so Joey could run all three races, in theory), then there are four possibilities for the second and third races, with the following probabilities:Joey runs both races: (1/12)(1/12) = 1/144Joey runs 2nd but not 3rd: (1/12)(11/12) = 11/144Joey runs 3rd but not 2nd: (1/12)(11/12) = 11/144Joey does not run: (11/12)(11/12) = 121/144If you want to answer the question "what is the probability Joey runs in at least one of the second and third races?" you would add the first three probabilities above (you would not subtract one of them, as you did) to get 23/144. It's more efficient, though, to just find the probability Joey does not run at all, which is 121/144, and subtract that from 1.varotkorn wrote:Q2. Since the original question is NON-replacement (it's like choosing 3 positions SIMULTANEOUSLY), it is mutually exclusive (meaning that Joey can't run second and third at the same time). Hence, the (1/12)*(1/12) part disappears?Yes, that's one way to look at it.varotkorn wrote:Q3. We can add together 1/12 + 1/12 ONLY when all the 12 members are DISTINCT, right? This approach cannot be generalized to, say, 4 red marbles and 8 blue marbles where there are repeats.Yes, that's true, because picking red second and picking red third are not mutually exclusive if you have several red marbles - both things can happen. But you could just answer this kind of question using the multiplication methods I used above (being sure to correctly account for whether the selections are made with or without replacement).

 Author: kornn [ 14 Jun 2020, 20:32 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te IanStewart wrote:varotkorn wrote:Q3. We can add together 1/12 + 1/12 ONLY when all the 12 members are DISTINCT, right? This approach cannot be generalized to, say, 4 red marbles and 8 blue marbles where there are repeats.Yes, that's true, because picking red second and picking red third are not mutually exclusive if you have several red marbles - both things can happen. But you could just answer this kind of question using the multiplication methods I used above (being sure to correctly account for whether the selections are made with or without replacement).Dear IanStewart,Modified Question: If there are 4 red marbles and 8 blue marbles, what is the probability that the 2nd or 3rd pick (WITHOUT REPLACEMENT) is red?I've tried both the methods. The result is amazing! (Not sure whether it is a coincidence though )I'm not sure about the highlighted part in Venn Diagram Method.P(R2nd) = P(R3rd) because regardless of the order we pick P(R Nth pick) = P(R 1st pick), right?Multiplication MethodTotal Outcomes = 12*11*10 = 1320Case 1: RRR = 4*3*2 = 24Case 2: BRR = 8*4*3 = 96Case 3: RBR = 96 (same as above)Case 4: RRB = 96 (same as above)Case 5: BBR = 8*7*4 = 224Case 6: BRB = 224 (same as above)Total desired outcomes = 760Probability = 760/1320 = 19/33Venn Diagram MethodP(R2nd OR R3rd) = P(R2nd) + P(R3rd) - P(R2nd AND R3d --> Case 1 and 2 above) = 4/12 + 4/12 - (24+96)/1320 = 19/33IMO, Venn Diagram Method is a bit quicker for me, although the highlighted parts are the trickiest.

 Author: IanStewart [ 15 Jun 2020, 01:10 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te varotkorn wrote:Modified Question: If there are 4 red marbles and 8 blue marbles, what is the probability that the 2nd or 3rd pick (WITHOUT REPLACEMENT) is red?The probability the second or third is red will be the same as the probability the first or second is red. It's easiest to work out the probability they are both blue, which is (8/12)(7/11) = 14/33 and subtract from 1, to get 19/33. I didn't check your work, but If you got the same answer, your solutions must be right - that wouldn't happen by coincidence.

 Author: kornn [ 15 Jun 2020, 01:37 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te IanStewart wrote:The probability the second or third is red will be the same as the probability the first or second is red.I've seen the above formula quite often, but I'm not sure what is the limitation for the formula.One quick question sir:Will the probability the THIRD or FIFTH or TENTH is red still be the same as the probability the FIRST or SECOND or THIRD is red as well?

 Author: IanStewart [ 15 Jun 2020, 05:14 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te varotkorn wrote:Will the probability the THIRD or FIFTH or TENTH is red still be the same as the probability the FIRST or SECOND or THIRD is red as well?If you have 4 red and 8 blue marbles, and you pick ten of them, then assuming you have no information about the first nine selections, the probability the tenth marble will be red is 4/12 = 1/3. The tenth selection is no more or less likely to be red than the first selection.

 Author: Nikhil30 [ 04 Jul 2021, 20:23 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te i tried with combination approach can someone plzz suggest where i went wrong??1. joye run second : 11c1*1*10c1/12c32. joye run third : 11c1*10c1*1/12c3is this wrong??i am getting a weird answer by this approach.plzz suggest.

 Author: IanStewart [ 05 Jul 2021, 02:54 ] Post subject: Re: A certain team has 12 members, including Joey. A three-member relay te Nikhil30 wrote:i tried with combination approach can someone plzz suggest where i went wrong??1. joye run second : 11c1*1*10c1/12c32. joye run third : 11c1*10c1*1/12c3is this wrong??i am getting a weird answer by this approach.plzz suggest.We're picking someone to run first, someone to run second, and someone to run third, so order matters (if we pick Joey to run first, that's different from picking Joey to run third). So it's not a combinations problem, and the total number of selections we can make is not "12C3"; it is 12*11*10. If in your solution, you change each "12C3" to 12*11*10 (and you can change "11C1" and "10C1" into 11 and 10 at the same time), then you'll have the correct probabilities for each case, and adding those together will give you the right answer.

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