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# Probability: Football

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Retired Moderator
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17 Feb 2009, 18:21
During the break of a football match the coach will make 3 substitutions. If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?

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17 Feb 2009, 19:16
During the break of a football match the coach will make 3 substitutions. If the team consists of 11 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted?

It could be with or without replacement but since the substitutions is made during the break so there is no replecement possible. so

= 9c3/11c3
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Director
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19 Feb 2009, 14:22
$$(9/11)*(8/10)*(7/9) = 28/55$$

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19 Feb 2009, 15:32

botirvoy wrote:
$$(9/11)*(8/10)*(7/9) = 28/55$$

P1=prob. of 1st non-forward sub=9/11
P2= prob. of 2st non-forward sub=8/10
P3= prob. of 3st non-forward sub=7/9

P=P1*P2*P3

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Retired Moderator
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20 Feb 2009, 07:07

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Manager
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08 Jul 2010, 11:41
Why can't we do it in this way?

The answers don't match..can some one tell me where i am going wrong.

1 - (probability of one of 2 forwards replaced + prob of both fwds getting replaced)
= 1 - (9C1 * 2C2 + 9C2 * 2C1)/11C3

Thanks

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08 Jul 2010, 11:56
As I have not read probability during my college, so I am not used to the formulae. I just do it conceptually & if dont find any way, I just make a guess & move forward. Here was my take.

Just consider this formation. I imagined that there are two red balls & nine white balls in a bag. I have to pick 3 balls such that not a single is red amongst them.

Red-Red-White-White-White-White-White-White-White-White-White

Probabilty of choosing white balls & not a single red ball will be:
$$\frac{9}{11} * \frac{8}{10} * \frac{7}{9} = \frac{28}{55}$$
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Re: Probability: Football   [#permalink] 08 Jul 2010, 11:56
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