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Coach John has to form a team of 4 players to be selected from a pool [#permalink]
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29 Jan 2015, 08:43
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Re: Coach John has to form a team of 4 players to be selected from a pool [#permalink]
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29 Jan 2015, 10:05
Bunuel wrote: Coach John has to form a team of 4 players to be selected from a pool of 6 players: Alan, Ben, Charles, Danny, Edward and Frank. John does not want Ben on the team but definitely wants Edward. Also, John does not want Alan and Charles on the same team. If John makes a random selection of 4 players, what is the probability that he will have a team that fulfills his criteria?
A. 1/30 B. 1/15 C. 2/15 D. 1/3 E. 2/3
Kudos for a correct solution. ans C... 2/15 since ben is out and edward is in, we are looking for remaining 3 players... danny and frank are also in with fourth seat between alan /charles... so total ways of selection of his desired team=2*4!.. total ways of selecting 4 out of 6=6*5*4*3=360 prob=2*24/360=2/15
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Re: Coach John has to form a team of 4 players to be selected from a pool [#permalink]
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02 Feb 2015, 03:59
Bunuel wrote: Coach John has to form a team of 4 players to be selected from a pool of 6 players: Alan, Ben, Charles, Danny, Edward and Frank. John does not want Ben on the team but definitely wants Edward. Also, John does not want Alan and Charles on the same team. If John makes a random selection of 4 players, what is the probability that he will have a team that fulfills his criteria?
A. 1/30 B. 1/15 C. 2/15 D. 1/3 E. 2/3
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:In a (random) probability problem, the probability of a favorable outcome is calculated as the number of favorable outcomes divided by the number of total outcomes, and often the number of total outcomes is easiest to calculate. Here, we have 6 items from which to choose 4, so that can be calculated using the Combinations formula of 6!/(4!2!), which equals 15. (Note, even if that's as far as you get you can eliminate A as the denominator is either 15 or a factor of 15 at this point) Then we need to determine how many favorable outcomes there are. The only teams that will work will include: Edward (who must be on the team for it to be favorable) Exactly one of Alan and Charles Danny Frank (note, because you can't use Ben and you can't use one of Alan/Charles, the other two spots MUST BE Danny and Frank to form a 4person team out of 6). This means that there are only two favorable teams. A, B, D, F and A, C, D, F. Therefore the numerator of the probability setup is 2, and the answer must be C.
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Coach John has to form a team of 4 players to be selected from a pool [#permalink]
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26 Sep 2015, 04:56
Bunuel wrote: Bunuel wrote: Coach John has to form a team of 4 players to be selected from a pool of 6 players: Alan, Ben, Charles, Danny, Edward and Frank. John does not want Ben on the team but definitely wants Edward. Also, John does not want Alan and Charles on the same team. If John makes a random selection of 4 players, what is the probability that he will have a team that fulfills his criteria?
A. 1/30 B. 1/15 C. 2/15 D. 1/3 E. 2/3
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:In a (random) probability problem, the probability of a favorable outcome is calculated as the number of favorable outcomes divided by the number of total outcomes, and often the number of total outcomes is easiest to calculate. Here, we have 6 items from which to choose 4, so that can be calculated using the Combinations formula of 6!/(4!2!), which equals 15. (Note, even if that's as far as you get you can eliminate A as the denominator is either 15 or a factor of 15 at this point) Then we need to determine how many favorable outcomes there are. The only teams that will work will include: Edward (who must be on the team for it to be favorable) Exactly one of Alan and Charles Danny Frank (note, because you can't use Ben and you can't use one of Alan/Charles, the other two spots MUST BE Danny and Frank to form a 4person team out of 6). This means that there are only two favourable teams. A, B, D, F and A, C, D, F. Therefore the numerator of the probability setup is 2, and the answer must be C. Hello Bunuel,
Can you explain why we have A and C in the same group even when the question says A and C should not be in the same team. Also why E is out when question says E must be in. Can we replace A or C with E?
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Re: Coach John has to form a team of 4 players to be selected from a pool [#permalink]
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04 Mar 2016, 19:19
ok, so this is a question in which we can actually list all the options... first thing, 6C4 = 6*5/2 = 15 ways to arrange without restrictions. thus, 15 is our denominator.
Ben can't be chosen..so we are left with only 5 players. we need E to be 100% in the team... we are left with 4 players, out of which A and C can't be together. We have only 2 possibilities: A, D, E, F or C, D, E, F
2 options  so 2 is our numerator
2/15 C



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Re: Coach John has to form a team of 4 players to be selected from a pool [#permalink]
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01 Apr 2016, 13:57
Learning4mU wrote: Bunuel wrote: Bunuel wrote: Coach John has to form a team of 4 players to be selected from a pool of 6 players: Alan, Ben, Charles, Danny, Edward and Frank. John does not want Ben on the team but definitely wants Edward. Also, John does not want Alan and Charles on the same team. If John makes a random selection of 4 players, what is the probability that he will have a team that fulfills his criteria?
A. 1/30 B. 1/15 C. 2/15 D. 1/3 E. 2/3
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:In a (random) probability problem, the probability of a favorable outcome is calculated as the number of favorable outcomes divided by the number of total outcomes, and often the number of total outcomes is easiest to calculate. Here, we have 6 items from which to choose 4, so that can be calculated using the Combinations formula of 6!/(4!2!), which equals 15. (Note, even if that's as far as you get you can eliminate A as the denominator is either 15 or a factor of 15 at this point) Then we need to determine how many favorable outcomes there are. The only teams that will work will include: Edward (who must be on the team for it to be favorable) Exactly one of Alan and Charles Danny Frank (note, because you can't use Ben and you can't use one of Alan/Charles, the other two spots MUST BE Danny and Frank to form a 4person team out of 6). This means that there are only two favourable teams. A, B, D, F and A, C, D, F. Therefore the numerator of the probability setup is 2, and the answer must be C. Hello Bunuel,
Can you explain why we have A and C in the same group even when the question says A and C should not be in the same team. Also why E is out when question says E must be in. Can we replace A or C with E?
it should be A D E F or C D E F. proably appears to be some typo error



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Re: Coach John has to form a team of 4 players to be selected from a pool [#permalink]
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Re: Coach John has to form a team of 4 players to be selected from a pool
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