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.

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:

A certain team has 12 members, including Joey. A three [#permalink]
25 Apr 2012, 09:59

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

68% (02:11) correct
32% (01:23) wrong based on 219 sessions

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. 1/1,320 B. 1/132 C. 1/110 D. 1/12 E. 1/6

Guys actually i have an official explanation of the right answer, but its a bit illogical for me

Explanation: 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 =1/6

but 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 third Could u please share with your thoughts on this?

Re: A certain team has 12 members, including Joey. A three [#permalink]
25 Apr 2012, 10:37

5

This post received KUDOS

Expert's post

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. 1/1,320 B. 1/132 C. 1/110 D. 1/12 E. 1/6

Guys actually i have an official explanation of the right answer, but its a bit illogical for me

Explanation: 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 =1/6

but 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 third Could 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...

Re: A certain team has 12 members, including Joey. A three [#permalink]
25 Apr 2012, 10:54

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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).

Re: A certain team has 12 members, including Joey. A three [#permalink]
26 Apr 2012, 12:05

Bunuel i 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

Re: A certain team has 12 members, including Joey. A three [#permalink]
26 Apr 2012, 12:54

Expert's post

Galiya wrote:

Bunuel i 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

Re: A certain team has 12 members, including Joey. A three [#permalink]
07 Jun 2013, 14:07

3

This post received KUDOS

Let s say that we have three spots to fill _ _ _ one for each position

Case 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} _________________

It is beyond a doubt that all our knowledge that begins with experience.

Probability of Joey getting selected at second or third place (P) = 1 - (Probability of Joey either not selected at all (P1) or Joey selected at first place (P2) )

P1 = Joey is not selected, first person is selected out of 11, second is selected out of 10 and third is selected out of 9/Total possbile outcomes = (11*10*9)/(12*11*10) P2 = Joey is selected at first, Second person is selected out of 11 and third is selected out of 10/Total possible outcomes = (1*11*10)/(12*11*10)

Re: A certain team has 12 members, including Joey. A three [#permalink]
31 Oct 2013, 21:32

1

This post received KUDOS

probability that Joey will be chosen to run second or third Means, 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

gmatclubot

Re: A certain team has 12 members, including Joey. A three
[#permalink]
31 Oct 2013, 21:32

It’s been a long time, since I posted. A busy schedule at office and the GMAT preparation, fully tied up with all my free hours. Anyways, now I’m back...

Ah yes. Funemployment. The time between when you quit your job and when you start your MBA. The promised land that many MBA applicants seek. The break that every...

It is that time of year again – time for Clear Admit’s annual Best of Blogging voting. Dating way back to the 2004-2005 application season, the Best of Blogging...