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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.
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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.
Show more
not sure if this is correct but i get 52/55
none of forwards substituted =
1 - all forwards picked
total possible = 11C3 = 11!/ 3!8! => 11*10*9 / 3*2*1 = 55
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?
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.
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.
Show more
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
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.