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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

Re: combinations - at least [#permalink]
27 Sep 2009, 11:13

4

This post received KUDOS

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

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.

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, 05:12

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, 13:03

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

Re: combinations - at least [#permalink]
26 May 2010, 16:16

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.

Re: A coach will make 3 substitutions. The team has 11 players, [#permalink]
22 Jan 2013, 03:03

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, 06:18

1

This post received KUDOS

Expert's post

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]
17 Mar 2014, 21:43

Expert's post

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.