7C3/9C5 = 35/126=5/18canuckclint wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
OA to come soon!
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baseball team probability
I have a question, since I just joined this site and I'm not good with probabilites and such. What do you folks mean by 9C5 and such? Please explain to me how 9C5 equates to 126.
9C5 means that its a combination. The formula for a combination is n!/k! (n-k!) Where n is the total number of possible items (in this case 9 players) and k is the total number of slots (in this case 5 spots on the team)silencz wrote:I have a question, since I just joined this site and I'm not good with probabilites and such. What do you folks mean by 9C5 and such? Please explain to me how 9C5 equates to 126.
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Is it okay to solve the problem this way:
He has to choose 5 players from a group of 9 players, so 5/9.
Then, with one less player to choose from, the probability of being chosen as the second player on the team is 4/8. (This is because the person who was selected to be the first player on the team can't also be selected as the second player on the team).
Finally, multiply the two probabilities:
(5/9)(4/8) = 20/72 = 5/18
Compared to some of the other explanations, this seems too simple. Is this method correct?
He has to choose 5 players from a group of 9 players, so 5/9.
Then, with one less player to choose from, the probability of being chosen as the second player on the team is 4/8. (This is because the person who was selected to be the first player on the team can't also be selected as the second player on the team).
Finally, multiply the two probabilities:
(5/9)(4/8) = 20/72 = 5/18
Compared to some of the other explanations, this seems too simple. Is this method correct?
John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
5 players to be choosen so first player can be choosen in 9 ways ,second player in 8 ways (as after selection of first player only 8 players are left) 3 rd player in 7 ways and 4 th player in 6 ways and fifth player can be choosen in 5 ways
so toal number of ways of choosing player 9*8*7*6*5
now we want John and Peter always so only 3 player needs to be choosen out of remaining 7 players
so again first palyer can be choosen in 7 ways second player in 6 ways 3 rd player in 5 ways so total number of ways of doing this 7*6*5
so probability=(7*6*5)/(9*8*7*6*5 ) =1/ 9*8=1/72
hope it helps
5 players to be choosen so first player can be choosen in 9 ways ,second player in 8 ways (as after selection of first player only 8 players are left) 3 rd player in 7 ways and 4 th player in 6 ways and fifth player can be choosen in 5 ways
so toal number of ways of choosing player 9*8*7*6*5
now we want John and Peter always so only 3 player needs to be choosen out of remaining 7 players
so again first palyer can be choosen in 7 ways second player in 6 ways 3 rd player in 5 ways so total number of ways of doing this 7*6*5
so probability=(7*6*5)/(9*8*7*6*5 ) =1/ 9*8=1/72
hope it helps
Ideation without execution is delusion
A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?
(A) 1/9
(B) 1/6
(C) 2/9
(D) 5/18
(E) 1/3
The OA is (D), 5/18.
Can someone please demonstrate the most efficient (fastest) way to obtain the correct answer? Thanks in advance!
[/list]
(A) 1/9
(B) 1/6
(C) 2/9
(D) 5/18
(E) 1/3
The OA is (D), 5/18.
Can someone please demonstrate the most efficient (fastest) way to obtain the correct answer? Thanks in advance!
[/list]-
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The word team should make you think combination.
The question is asking for the following fraction:
(total number of good teams)/(total number of possible teams)
That's why it's best to treat this as a combination question, determining each part of the fraction separately.
To determine the total number of possible teams, we need to determine how many combinations of 5 can be made from 9 choices:
(9*8*7*6*5)/(1*2*3*4*5) = 126 total possible teams
A good team will consist of John and Peter put together with 3 other players. Since we know that John and Peter have to be included in a good team, our only concern is the number of ways we can choose the 3 other players from the 7 we have left.
So we need to determine how many combinations of 3 can be made from 7 choices:
(7*6*5)/(1*2*3) = 35 good teams.
So (good teams)/(total teams) = 35/126 = 5/18.
The correct answer is D.
The question is asking for the following fraction:
(total number of good teams)/(total number of possible teams)
That's why it's best to treat this as a combination question, determining each part of the fraction separately.
To determine the total number of possible teams, we need to determine how many combinations of 5 can be made from 9 choices:
(9*8*7*6*5)/(1*2*3*4*5) = 126 total possible teams
A good team will consist of John and Peter put together with 3 other players. Since we know that John and Peter have to be included in a good team, our only concern is the number of ways we can choose the 3 other players from the 7 we have left.
So we need to determine how many combinations of 3 can be made from 7 choices:
(7*6*5)/(1*2*3) = 35 good teams.
So (good teams)/(total teams) = 35/126 = 5/18.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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