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

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.
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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 = \(\frac{2*10*11}{10*11*12}=\frac{1}{6}\)

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

Hope it helps.
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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

Go through the questions here: search.php?search_id=tag&tag_id=54

There should be a lot of the questions of that type.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE


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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.

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

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

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
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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.

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).

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.

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.

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?

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.

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.

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.