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# A certain team has 12 members, including Joey. A three

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Manager
Joined: 16 Jan 2011
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A certain team has 12 members, including Joey. A three  [#permalink]

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25 Apr 2012, 10:59
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74% (01:09) correct 26% (01:33) wrong based on 757 sessions

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

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25 Apr 2012, 11:37
19
11
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

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.

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...

Hope it's clear.
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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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07 Jun 2013, 15:07
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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 = $$\frac{2*10*11}{10*11*12}=\frac{1}{6}$$
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##### General Discussion
Manager
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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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25 Apr 2012, 11:45
1
if rephrase the question "What is the probability that Joey will run second or third?", will we get the probability 1/11+1/10?
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Posts: 49303
Re: A certain team has 12 members, including Joey. A three  [#permalink]

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25 Apr 2012, 11:54
2
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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:
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
a-certain-class-consists-of-8-students-including-kim-127730.html

Hope it helps.
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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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26 Apr 2012, 13:05
Bunuel
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
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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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26 Apr 2012, 13:54
1
1
Galiya wrote:
Bunuel
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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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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07 Jun 2013, 07:03
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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

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

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

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07 Jun 2013, 08:15
2
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)

$$1 - \frac{1}{12} - (\frac{11}{12} * \frac{10}{11} * \frac{9}{10})$$

$$1 - \frac{1}{12} - \frac{9}{12}$$

$$1 - \frac{10}{12}$$

$$\frac{2}{12} = \frac{1}{6}$$

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

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07 Jun 2013, 08:20
lchen wrote:
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)

$$1 - \frac{1}{12} - (\frac{11}{12} * \frac{10}{11} * \frac{9}{10})$$

$$1 - \frac{1}{12} - \frac{9}{12}$$

$$1 - \frac{10}{12}$$

$$\frac{2}{12} = \frac{1}{6}$$

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.

Sure, 1/6 is E, not A. Edited. Thank you.
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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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23 Oct 2013, 18:33
Bunuel wrote:
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

Alternative Solution:

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)

P1 + P2 = (11*10*9 + 1*11*10)/(12*11*10)
= 10/12

P = 1 - (P1+P2)
= 1- (10/12)
= 2/12
=1/6
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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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31 Oct 2013, 22:32
1
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
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A certain team has 12 members, including Joey. A three  [#permalink]

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26 Sep 2015, 11:25
1
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

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

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06 Apr 2017, 04:34
Hello,

I solved it this way:

(Probability of Joey not getting selected first * Probability of Joey getting selected second) + (Probability of Joey not getting selected first * Probability of Joey not getting selected second * Probability of Joey getting selected third)

= (11/12 * 1/11) + (11/12 * 10/11 * 1/10)

= 1/12 + 1/12

=2/12 = 1/6.

Please let me know if this method can be used for all problems of this type.

Thank you.
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Re: A certain team has 12 members, including Joey. A three  [#permalink]

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25 Jul 2018, 03:03
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Re: A certain team has 12 members, including Joey. A three &nbs [#permalink] 25 Jul 2018, 03:03
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