A certain team has 12 members, including Joey. A three-member relay te

Question

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

[

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?

Bunuel · Accepted Answer

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.

Zarrolou · Answer

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}

galiya · Answer

if rephrase the question "What is the probability that Joey  will run  second or third?", will we get the probability 1/11+1/10?

Bunuel · Answer

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.html https://gmatclub.com/forum/a-medical-res ... 27396.html https://gmatclub.com/forum/a-certain-cla ... 27730.html Hope it helps.

galiya · Answer

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

Bunuel · Answer

Go through the questions here: search.php?search_id=tag&tag_id=54 There should be a lot of the questions of that type.

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

lchen · Answer

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 E

Explanation: 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.

vivmechster · Answer

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

koreye · Answer

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 =  1/12
 
 Case 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 =  1/12

Now 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 = 1/6

Answer: E

ScottTargetTestPrep · Answer

The probability Joey will be chosen to run second is:

11/12 x 1/11 x 10/10 = 1/12

The probability Joey will be chosen to run third is:

11/12 x 10/11 x 1/10 = 1/12

Thus, the probability that he will be chosen to run second or third is:

1/12 + 1/12 = 2/12 = 1/6

Answer: E

kornn · Answer

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.

IanStewart · Answer

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) = 1/144
Joey runs 2nd but not 3rd: (1/12)(11/12) = 11/144
Joey runs 3rd but not 2nd: (1/12)(11/12) = 11/144
Joey does not run: (11/12)(11/12) = 121/144

If 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.

Yes, that's one way to look at it.

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).

kornn · Answer

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 Method Total Outcomes = 12*11*10 = 1320 Case 1: RRR = 4*3*2 = 24 Case 2: BRR = 8*4*3 = 96 Case 3: RBR = 96 (same as above) Case 4: RRB = 96 (same as above) Case 5: BBR = 8*7*4 = 224 Case 6: BRB = 224 (same as above) Total desired outcomes = 760 Probability = 760/1320 = 19/33 Venn Diagram Method P(R2nd OR R3rd) = P(R2nd) + P(R3rd) - P(R2nd AND R3d --> Case 1 and 2 above) = 4/12 + 4/12 - (24+96)/1320 = 19/33 IMO, Venn Diagram Method is a bit quicker for me, although the highlighted parts are the trickiest.

IanStewart · Answer

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.

kornn · Answer

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?

IanStewart · Answer

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.

Nikhil30 · Answer

i tried with combination approach can someone plzz suggest where i went wrong??
1. joye run second : 11c1*1*10c1/12c3
2. joye run third : 11c1*10c1*1/12c3

is this wrong??
i am getting a weird answer by this approach.

plzz suggest.

IanStewart · Answer

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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