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A coach will make 3 substitutions. The team has 11 players, [#permalink]
 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?
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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
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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
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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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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
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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
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Re: combinations - at least [#permalink]
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
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Re: combinations - at least [#permalink]
08 Feb 2010, 00:03
nice one srinivas.
+1
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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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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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Re: combinations - at least [#permalink]
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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Re: combinations - at least [#permalink]
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.
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Re: combinations - at least [#permalink]
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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Re: combinations - at least [#permalink]
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?
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Re: A coach will make 3 substitutions. The team has 11 players, [#permalink]
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]
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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Re: A coach will make 3 substitutions. The team has 11 players, [#permalink]
22 Jan 2013, 07:30
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Re: A coach will make 3 substitutions. The team has 11 players,
[#permalink]
22 Jan 2013, 07:30
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