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bmwhype2
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thats what im thinking, 9!/3!6!.................................don't know if it's that easy tho
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bmwhype2
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bmwhype2
A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards. What is the probability that none of the forwards will be substituted?


Please explain your logic.



desired scenario / total possibilities


9C3 / 11C3
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There are 9*8*7 ways to arrange 3 substitutes that are not forwards. There are 11*10*9 ways to arrange all the players including substitutes. Hence the probability of selecting only players that are not forward is 9*8*7/11*10*9 = 28/55. Someone correct this if wrong.
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bmwhype2
A coach will make 3 substitutions. The team has 11 players, among which there are 2 forwards. What is the probability that none of the forwards will be substituted?


Please explain your logic.


Prob(3 substituted but none is forward) = Prob(Pick 1st) x Prob(Pick 2nd) x Prob(Pick 3rd)
= (9/11)x(8/10)x(7/9) = 28/55

Or
Prob = N(Events)/N(Sample Space)
= (9C3)/(11C3) = 28/55



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