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A coach will make 3 substitutions. The team has 11 players,

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CEO
Joined: 21 Jan 2007
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A coach will make 3 substitutions. The team has 11 players, [#permalink]

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19 Dec 2007, 13:58
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A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards.

a) What is the probability that at least one of the forwards will be substituted?

b) What is the probability that at least two of the forwards will be substituted?
CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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19 Dec 2007, 14:25
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The total number: N=9P3=990

a) M=1-9P3/N=1-9*8*7/990=1-56/110=54/110

b) R=9P1*3P3/N=9*3*2/990=6/110
CEO
Joined: 21 Jan 2007
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19 Dec 2007, 14:26
walker wrote:
The total number: N=9P3=990

a) M=1-9P3/N=1-9*8*7/990=1-56/110=54/110

b) R=9P1*3P3/N=9*3*2/990=6/110

can you show me via combinations rather than perm
CEO
Joined: 17 Nov 2007
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19 Dec 2007, 14:33
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bmwhype2 wrote:
can you show me via combinations rather than perm

The problem is sensitive to the order of substitutions: (forward,forward,half-back) and (half-back,forward,forward) are different substitutions.

Therefore, we should use permutation formula rather than combination formula.

You can change all nPm on nCm*m!
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19 Dec 2007, 14:47
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A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards.

a) What is the probability that at least one of the forwards will be substituted?

P(a) = 1 - Probability that none of the forwards will be substituted
= 1 - P(N)

P(N) = 9/11 x 8/10 x 7/9 = 56/110 = 28/55

So the answer is = 1 - 28/55 = = 54/110 = 27/55

b) What is the probability that at least two of the forwards will be substituted?

P(b) = P(a) - P(M)
P(M) = probability that only one will be selected = 2/11
27/55 - 2/11 = 27/55 - 10/55 = 17/55
CEO
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23 Dec 2007, 02:21
Total number: 11P3=990

1. none forward: p0=9P3/11P3=9!*8!/(6!*11!)=56/110

2. one forward: p1=2C1*9C2*3P3/11P3=2*9!*3!*8!/(7!*2!*11!)=48/110

3. two forwards: p2=2C2*9C1*3P3/11P3=1*9*6/990=6/110

4. at least one forward: p12=1-p0=1-56/110=54/110 or
p12=p1+p2=48/110+6/110=54/110
Manager
Joined: 27 Oct 2008
Posts: 185
Re: combinations - at least [#permalink]

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27 Sep 2009, 12:13
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A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards.

a) What is the probability that at least one of the forwards will be substituted?

b) What is the probability that at least two of the forwards will be substituted?

Soln:
a) What is the probability that at least one of the forwards will be substituted?
= 1 - Probability that none of the forwards will be substituted
= 1 - (9/11 * 8 /10 * 7/9)
= 1 - 28/55
= 27/55

b) What is the probability that at least two of the forwards will be substituted?
= (2/11 * 1/10 * 9/9) * 3
= 3/55
Manager
Joined: 17 Jan 2010
Posts: 147
Concentration: General Management, Strategy
GPA: 3.78
WE: Engineering (Manufacturing)

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12 Feb 2010, 09:46
walker wrote:
Total number: 11P3=990

1. none forward: p0=9P3/11P3=9!*8!/(6!*11!)=56/110

2. one forward: p1=2C1*9C2*3P3/11P3=2*9!*3!*8!/(7!*2!*11!)=48/110

3. two forwards: p2=2C2*9C1*3P3/11P3=1*9*6/990=6/110

4. at least one forward: p12=1-p0=1-56/110=54/110 or
p12=p1+p2=48/110+6/110=54/110

Could you please explain why you think it should be permutation, although in this case the answer is the same as you will have 3! in both numerator and denominator. The way I look at that is that it does not matter what order the players were chosen. The question states only that 3 players was chosen at random.
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12 Feb 2010, 10:25
alexBLR wrote:
Could you please explain why you think it should be permutation, although in this case the answer is the same as you will have 3! in both numerator and denominator. The way I look at that is that it does not matter what order the players were chosen. The question states only that 3 players was chosen at random.

Most probability/counting problems have a few equally right solutions. So, we can use any way.
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Senior Manager
Joined: 22 Dec 2009
Posts: 359
Re: combinations - at least [#permalink]

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16 Feb 2010, 06:12
bmwhype2 wrote:
A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards.

a) What is the probability that at least one of the forwards will be substituted?

b) What is the probability that at least two of the forwards will be substituted?

Somehow this question isn't very clear to me...

Is the coach going to make 3 substitutions at once.... as in if the 3 people from the playing 11 be replaced in one go...

or

would the substitutions be made one after the other..

As per the answer would differ accordingly.

Please comment.... might be my understanding is wrong...!
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Intern
Joined: 19 Apr 2010
Posts: 1
Re: combinations - at least [#permalink]

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02 May 2010, 14:03
srivas wrote:
A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards.

a) What is the probability that at least one of the forwards will be substituted?

b) What is the probability that at least two of the forwards will be substituted?

Soln:
a) What is the probability that at least one of the forwards will be substituted?
= 1 - Probability that none of the forwards will be substituted
= 1 - (9/11 * 8 /10 * 7/9)
= 1 - 28/55
= 27/55

b) What is the probability that at least two of the forwards will be substituted?
= (2/11 * 1/10 * 9/9) * 3
= 3/55

I think both of these are correct.
Manager
Joined: 15 Mar 2010
Posts: 99
Re: combinations - at least [#permalink]

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02 May 2010, 21:16
bmwhype2 wrote:
A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards.

a) What is the probability that at least one of the forwards will be substituted?

b) What is the probability that at least two of the forwards will be substituted?

My take:

(a) prob of choosing none forwards = 9C3/11C3
prob of choosing atleast one forward = 1- 9C3/11C3 = 27/55

(b) prob of choosing atleast 2 forwards = prob of choosing 2 forwards =2C2*9C1 / 11C3 = 3/55
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Intern
Joined: 24 May 2010
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Re: combinations - at least [#permalink]

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26 May 2010, 17:16
srivas wrote:
A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards.

a) What is the probability that at least one of the forwards will be substituted?

b) What is the probability that at least two of the forwards will be substituted?

Soln:
a) What is the probability that at least one of the forwards will be substituted?
= 1 - Probability that none of the forwards will be substituted
= 1 - (9/11 * 8 /10 * 7/9)
= 1 - 28/55
= 27/55

b) What is the probability that at least two of the forwards will be substituted?
= (2/11 * 1/10 * 9/9) * 3
= 3/55

Maybe I am not reading this question correctly or don't quite understand it but regarding question 1, wouldn't the probability be:
=1- (9/11 * 9/11 * 9/11)

Everytime the coach makes a substitution, a new player is on the field but is there any reason that the second sub can't enter the game for the first sub? If not, there should still be 11 substitutable players on the field of which 9 aren't forwards, hence the second 9/11.

Using this theory, the probability would be

= 1 - 729/1331

What am I missing?
Manager
Joined: 07 Feb 2011
Posts: 105
Re: A coach will make 3 substitutions. The team has 11 players, [#permalink]

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22 Jan 2013, 04:03
Confused at how order matters in this problem. I could solve it using combinations but never would have thought to use permutations here
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Re: A coach will make 3 substitutions. The team has 11 players, [#permalink]

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22 Jan 2013, 07:18
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manimgoindowndown wrote:
Confused at how order matters in this problem. I could solve it using combinations but never would have thought to use permutations here

The order does not matter if all positions on the team are the same. Say, there are 11 players in the team. You need to substitute 3 of them with 3 other players. You CHOOSE any 3 players out of the 11 and put the other 3 in their place. This is a combinations problem.

In this question, there are 2 forwards and one other position.
Say if there are 11 players P1, P2 to P11. P1 and P2 are forwards.
You have 3 other players A, B and C.
When you substitute, it is different if A and B are made forwards from the case where B and C are forwards. This problem involves permutation too.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 39724 Re: A coach will make 3 substitutions. The team has 11 players, [#permalink] Show Tags 22 Jan 2013, 07:30 Similar questions to practice: a-coach-is-filling-out-the-starting-lineup-for-his-indoor-85800.html Hope it helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7447 Location: Pune, India Re: A coach will make 3 substitutions. The team has 11 players, [#permalink] Show Tags 17 Mar 2014, 22:43 bmwhype2 wrote: A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards. a) What is the probability that at least one of the forwards will be substituted? b) What is the probability that at least two of the forwards will be substituted? Responding to a pm: There is one good thing about probability: you can often ignore the order even if it does matter as long as you ignore it in numerator as well as denominator. Check this post for more: http://www.veritasprep.com/blog/2013/08 ... er-matter/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A coach will make 3 substitutions. The team has 11 players,   [#permalink] 17 Mar 2014, 22:43
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