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John and Peter are among the nine pl ayers 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?
- a)1/9
- b)1/6
- c)2/9
- d)5/18
- e)1/3
Correct answer is option 'D'. Can you explain this answer?
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John and Peter are among the nine pl ayers a basketball coach can choo...
First we must find the total number of 5 member teams, with or without John and Peter. We can solve this using an anagram model in which each of the 9 players (A – I) is assigned either a Y (for being chosen) or an N (for not being chosen):
It is the various arrangements of Y’s and N’s above that would yield all of the different combinations, so we can find the number of possible teams here by considering how many anagrams of YYYYYNNNN exist:
(because there are 9! ways to order 9 objects) (because the 5Y's and 4N's are identical)
So there are 126 possible teams of 5. Since the question asks for the probability of choosing a team that includes John and Peter, we need to determine how many of the 126 include John and Peter. If we reserve two of the 5 spots on a team for John and Peter, there will be 3 spots left, which must be filled by 3 of the remaining 7 players (remember that John and Peter were already selected). Therefore the number of teams including John and Peter will be equal to the number of 3-player teams that can be formed from a 7-player pool. We can approach the problem as we did above:
The number of possible YYYNNNN anagrams is:
Since 35 of the total possible 126 teams include John and Peter, the probability of selecting a team with both John and Peter is 35/126 or 5/18.
It is the various arrangements of Y’s and N’s above that would yield all of the different combinations, so we can find the number of possible teams here by considering how many anagrams of YYYYYNNNN exist:
(because there are 9! ways to order 9 objects) (because the 5Y's and 4N's are identical)
So there are 126 possible teams of 5. Since the question asks for the probability of choosing a team that includes John and Peter, we need to determine how many of the 126 include John and Peter. If we reserve two of the 5 spots on a team for John and Peter, there will be 3 spots left, which must be filled by 3 of the remaining 7 players (remember that John and Peter were already selected). Therefore the number of teams including John and Peter will be equal to the number of 3-player teams that can be formed from a 7-player pool. We can approach the problem as we did above:
The number of possible YYYNNNN anagrams is:
Since 35 of the total possible 126 teams include John and Peter, the probability of selecting a team with both John and Peter is 35/126 or 5/18.
Most Upvoted Answer
John and Peter are among the nine pl ayers a basketball coach can choo...
Calculation:
Total number of ways to choose 5 players out of 9:
- This can be calculated using the combination formula C(n, r) = n! / (r!(n-r)!), where n is the total number of players and r is the number of players to be chosen.
- In this case, n = 9 (total players) and r = 5 (players to be chosen).
- Therefore, the total number of ways to choose 5 players out of 9 is C(9, 5) = 9! / (5! * 4!) = 126 ways.
Number of ways to choose a team that includes John and Peter:
- Since John and Peter are already included in the team, we need to choose 3 players from the remaining 7 players.
- This can be calculated using the combination formula C(7, 3) = 7! / (3! * 4!) = 35 ways.
Probability of choosing a team that includes John and Peter:
- Probability = Number of favorable outcomes / Total number of outcomes
- Probability = 35 (number of ways to choose a team with John and Peter) / 126 (total number of ways to choose 5 players out of 9)
- Probability = 35/126 = 5/18
Therefore, the probability of choosing a team that includes John and Peter is 5/18.
Total number of ways to choose 5 players out of 9:
- This can be calculated using the combination formula C(n, r) = n! / (r!(n-r)!), where n is the total number of players and r is the number of players to be chosen.
- In this case, n = 9 (total players) and r = 5 (players to be chosen).
- Therefore, the total number of ways to choose 5 players out of 9 is C(9, 5) = 9! / (5! * 4!) = 126 ways.
Number of ways to choose a team that includes John and Peter:
- Since John and Peter are already included in the team, we need to choose 3 players from the remaining 7 players.
- This can be calculated using the combination formula C(7, 3) = 7! / (3! * 4!) = 35 ways.
Probability of choosing a team that includes John and Peter:
- Probability = Number of favorable outcomes / Total number of outcomes
- Probability = 35 (number of ways to choose a team with John and Peter) / 126 (total number of ways to choose 5 players out of 9)
- Probability = 35/126 = 5/18
Therefore, the probability of choosing a team that includes John and Peter is 5/18.
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John and Peter are among the nine pl ayers 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?a)1/9b)1/6c)2/9d)5/18e)1/3Correct answer is option 'D'. Can you explain this answer?
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