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Re: John and Peter are among the nine players a basketball coach [#permalink]
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11 Mar 2014, 08:07
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CCMBA wrote: VeritasPrepKarishma wrote: CCMBA wrote: I did the lengthier solution, but I get something different.
1[Pr(Neither) + Pr(J, not P) + Pr(P, not J)]
1  [3/18+ 2/18]
13/18
What did I do wrong?
Ex: Pr(Neither) = 7/9*6/8...*3/5 = 1/6 P(Both not selected) = (4/9) * (3/8) = 1/6 (The probability of John being among the 4 leftover out of 9 people is 4/9 and probability of Peter being among the leftover 3 is 3/8) P(Peter selected, John is not) = (5/9)*(4/8) = 5/18 (Probability of Peter being among 5 selected is 5/9 and John being among the leftover 4 is 4/8) P(Peter not selected, John is selected) = (5/9)*(4/8) = 5/18 (Probability of John being among 5 selected is 5/9 and Peter being among the leftover 4 is 4/8) 1  (1/6 + 5/18 + 5/18) = 5/18 Hi Karishma, Thank you for your response. There are still 2 things I do not understand. The first is the fractions you use. Are we taking as given that the first 3 spots are filled? So for Pr(Neither), we're only considering the last spot, and we want to ignore J and P? Second, where is the flaw in the method I used? Pr(Neither) = 7/9 * 6/8 * 5/7 * 4/6 * 3/5 = 1/6 Pr(J, not P) = Pr(P, not J) = 1/9 * 7/8 * 6/7 * 5/6 * 4/5 = 1/18 Pr(J) is 1/9. Then we are only considering 7 of the remaining 8 people, then 6 of 7, etc. Thank you. go for easy solution ! with combinations or permatuation ! :D 2c2*7c3/9c5 or 5!/3!*1/9*1/8*1*1*1
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Re: John and Peter are among the nine players a basketball coach [#permalink]
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11 Mar 2014, 23:34
CCMBA wrote: VeritasPrepKarishma wrote: CCMBA wrote: I did the lengthier solution, but I get something different.
1[Pr(Neither) + Pr(J, not P) + Pr(P, not J)]
1  [3/18+ 2/18]
13/18
What did I do wrong?
Ex: Pr(Neither) = 7/9*6/8...*3/5 = 1/6 P(Both not selected) = (4/9) * (3/8) = 1/6 (The probability of John being among the 4 leftover out of 9 people is 4/9 and probability of Peter being among the leftover 3 is 3/8) P(Peter selected, John is not) = (5/9)*(4/8) = 5/18 (Probability of Peter being among 5 selected is 5/9 and John being among the leftover 4 is 4/8) P(Peter not selected, John is selected) = (5/9)*(4/8) = 5/18 (Probability of John being among 5 selected is 5/9 and Peter being among the leftover 4 is 4/8) 1  (1/6 + 5/18 + 5/18) = 5/18 Hi Karishma, Thank you for your response. There are still 2 things I do not understand. The first is the fractions you use. Are we taking as given that the first 3 spots are filled? So for Pr(Neither), we're only considering the last spot, and we want to ignore J and P? Second, where is the flaw in the method I used? Pr(Neither) = 7/9 * 6/8 * 5/7 * 4/6 * 3/5 = 1/6 Pr(J, not P) = Pr(P, not J) = 1/9 * 7/8 * 6/7 * 5/6 * 4/5 = 1/18 Pr(J) is 1/9. Then we are only considering 7 of the remaining 8 people, then 6 of 7, etc. Thank you. The probability of selecting J is 5/9 since 5 people have to be selected and he can be any one of the 5. If you use this, you will get the correct answer. Though this is a very round about way of solving the question. Check out the solutions given on page 1.
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Re: John and Peter are among the nine players a basketball coach [#permalink]
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13 Mar 2014, 09:17
CCMBA wrote: John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
I did the lengthier solution, but I get something different.
1[Pr(Neither) + Pr(J, not P) + Pr(P, not J)]
1  [3/18+ 2/18]
13/18
What did I do wrong?
Ex: Pr(Neither) = 7/9*6/8...*3/5 = 1/6 P = 1  (P(team with John and not Peter) + P(team with Peter but not John) + P(team without any of them)). \(P = 1  (\frac{1}{9}*\frac{7}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}*\frac{5!}{4!} + \frac{1}{9}*\frac{7}{8}*\frac{6}{7}*\frac{5}{6}*\frac{4}{5}*\frac{5!}{4!} + \frac{7}{9}*\frac{6}{8}*\frac{5}{7}*\frac{4}{6}*\frac{3}{5})=\frac{5}{18}\). We need to multiply first two cases by 5!/4! because JAAAA (John, Any but Peter, Any but Peter, Any but Peter, Any but Peter) and PAAAA (Peter, Any but John, Any but John, Any but John, Any but John) can occur in 5 ways. Hope it helps.
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Re: PS probability teams [#permalink]
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14 Apr 2014, 14:05
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tochiru wrote: How can we solve this in probability approach. I thought the answer would be 1/9 x 1/8 x 3/7. I have a different approach: The first one has 5 places out of 9, so 5/9 Then, the other has 4 places left, among 8, so 4/8 One thing AND the other must occur, so he multiply 5/9 * 1/2 = 5/18



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Re: John and Peter are among the nine players a basketball coach [#permalink]
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14 Jun 2015, 04:07
Bunuel wrote: tochiru wrote: How can we solve this in probability approach. I thought the answer would be 1/9 x 1/8 x 3/7. John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer 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 Probability approach:1/9 choosing John (J); 1/8 choosing Peter (P); 7/7=1 choosing any (A) for the third player; 6/6=1 choosing any (A) for the fourth player; 5/5=1 choosing any (A) for the fifth player. But scenario JPAAA can occur in \(\frac{5!}{3!}=20\) # of ways (JAPAA, JAAPA, AAAPJ, ... basically the # of permutations of the letters JPAAA, which is \(\frac{5!}{3!}=20\)). So \(P=\frac{5!}{3!}*\frac{1}{9}*\frac{1}{8}*1*1*1=\frac{5}{18}\). Combinatorics approach:P=favorable outcomes/total # of outcomes > \(P=\frac{C^2_2*C^3_7}{C^5_9}=\frac{5}{18}\). \(C^2_2=1\)  # of ways to choose Peter and John out of Peter and John, basically 1 way; \(C^3_7=35\)  # of ways to choose 3 other players out of 7 players left (without Peter and John); \(C^5_9=126\)  total # of ways to choose 5 players out of 9. Answer: D. Hope it helps. Hi Bunuel, I have a strange question here. We say the probability to pick the first player John is 1/9 ,Peter is 1/8 then why don't we say the third player will be 1/7 and so on.I am unable to differentiate between picking one player in 1/7 and picking any in 7/7? In picking any player,we are also picking just one player,then how is it different? Please spare some time and explain.It'll be of great help.



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Re: John and Peter are among the nine players a basketball coach [#permalink]
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14 Jun 2015, 12:38
davesinger786 wrote: Bunuel wrote: tochiru wrote: How can we solve this in probability approach. I thought the answer would be 1/9 x 1/8 x 3/7. John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer 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 Probability approach:1/9 choosing John (J); 1/8 choosing Peter (P); 7/7=1 choosing any (A) for the third player; 6/6=1 choosing any (A) for the fourth player; 5/5=1 choosing any (A) for the fifth player. But scenario JPAAA can occur in \(\frac{5!}{3!}=20\) # of ways (JAPAA, JAAPA, AAAPJ, ... basically the # of permutations of the letters JPAAA, which is \(\frac{5!}{3!}=20\)). So \(P=\frac{5!}{3!}*\frac{1}{9}*\frac{1}{8}*1*1*1=\frac{5}{18}\). Combinatorics approach:P=favorable outcomes/total # of outcomes > \(P=\frac{C^2_2*C^3_7}{C^5_9}=\frac{5}{18}\). \(C^2_2=1\)  # of ways to choose Peter and John out of Peter and John, basically 1 way; \(C^3_7=35\)  # of ways to choose 3 other players out of 7 players left (without Peter and John); \(C^5_9=126\)  total # of ways to choose 5 players out of 9. Answer: D. Hope it helps. Hi Bunuel, I have a strange question here. We say the probability to pick the first player John is 1/9 ,Peter is 1/8 then why don't we say the third player will be 1/7 and so on.I am unable to differentiate between picking one player in 1/7 and picking any in 7/7? In picking any player,we are also picking just one player,then how is it different? Please spare some time and explain.It'll be of great help. We want specifically John and Peter, but the remaining 3 players can be any from 7.
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Re: John and Peter are among the nine players a basketball coach [#permalink]
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14 Jun 2015, 19:37
I see. Alright,thanks for the clarification,Bunuel.



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John and Peter are among the nine players a basketball coach [#permalink]
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18 Jun 2015, 01:53
riteshgmat wrote: young_gun wrote: John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer 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 Total number of outcomes = 9*8*7*6*5/ 5*4*3*2*1 = 126 Total number of favorable outcomes including John and Peter = 7*6*5/3*2*1 = 35 Probability = 35/126 = 5 / 18 Hello, Could you explain how you calculated the outcomes: As I understand from your "remaining nubers", for the total number of outcomes you did: 9!/5!4!. Then you cancel out some of the numbers and end up with 504/4, which is 126. This is fine so far. What I cannot understand how you calculated the number of favorable outcomes. Could you also provide the whole ! you used? Thank you.



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Re: John and Peter are among the nine players a basketball coach [#permalink]
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27 Jun 2015, 04:12
Bunuel wrote: tochiru wrote: How can we solve this in probability approach. I thought the answer would be 1/9 x 1/8 x 3/7. John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer 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 Probability approach:1/9 choosing John (J); 1/8 choosing Peter (P); 7/7=1 choosing any (A) for the third player; 6/6=1 choosing any (A) for the fourth player; 5/5=1 choosing any (A) for the fifth player. But scenario JPAAA can occur in \(\frac{5!}{3!}=20\) # of ways (JAPAA, JAAPA, AAAPJ, ... basically the # of permutations of the letters JPAAA, which is \(\frac{5!}{3!}=20\)). So \(P=\frac{5!}{3!}*\frac{1}{9}*\frac{1}{8}*1*1*1=\frac{5}{18}\). Combinatorics approach:P=favorable outcomes/total # of outcomes > \(P=\frac{C^2_2*C^3_7}{C^5_9}=\frac{5}{18}\). \(C^2_2=1\)  # of ways to choose Peter and John out of Peter and John, basically 1 way; \(C^3_7=35\)  # of ways to choose 3 other players out of 7 players left (without Peter and John); \(C^5_9=126\)  total # of ways to choose 5 players out of 9. Answer: D. Hope it helps. I applied combinatorics approach but instead of 2C2 I used 9C2  ( # of ways to choose Peter and John out of Peter and John) and got messed.
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Re: John and Peter are among the nine players a basketball coach [#permalink]
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14 May 2016, 12:46
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Bunuel: Could you please tell me where i am wrong with my approach I wanna calculate the probability of selecting a team of 5 with out John and Peter and then subtract it with 1 Without John and Peter there are 7 members. So I can select 5 out of 7 in 7C5 ways. which is nothing but 21 Total number of ways to select 5 out of 9 are 9C5 which is nothing but 126 So probability of a team without John and Peter=21/126=1/6 Hence total probability with John and Peter=11/6=5/6 . Wrong answer Can anyone help



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Re: John and Peter are among the nine players a basketball coach [#permalink]
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22 Mar 2018, 15:49
young_gun wrote: John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer 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 The number of ways of choosing 5 players from 9 is 9C5 = 9!/[5!(9  5)!] = 9!/(5!4!) = (9 x 8 x 7 x 6 )/( 4 x 3 x 2) = 3 x 7 x 6 = 126. If John and Peter are already chosen for the team, then only 3 additional players must be chosen for the fiveplayer team. The number of ways of choosing the 3 additional players from the remaining 7 players is 7C3 = 7!/[3!(7  3)!] = 7!/(3!4!) = (7 x 6 x 5)/(3 x 2) = 7 x 5 = 35. Thus the probability that John and Peter will be chosen for the team is 35/126 = 5/18. Answer: D
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Re: John and Peter are among the nine players a basketball coach [#permalink]
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31 Mar 2018, 03:04
Hi Bunuel and moderators / anyone who's keen to help me out, Supposed that the coach decides to split the team of 9 players into 2 subteams (consisting of at least 2 person per subteam) for a training camp, how many possible subteams can be formed such that John and Peter are in the same subteam for the training camp? Would this be the same approach as: possible outcomes for 2 vs 7 person subteams and finding number of possibilities for each subteam with John & Peter, then repeat for 3 vs 6, 4 vs 5, 5 vs 4 etc.? I'm trying a different variant of the same question for practice thoroughness, but I'm stuck as to how to start counting the possibilities. Should I start by putting John in one of the subteams first, then count the probability of picking Peter, then others etc. for the different combinations of subteams? Would like to seek your advice on how to do this effective and efficiently. Regards, Kim




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