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John and Peter are among the nine players a basketball coach
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31 Jul 2008, 15:56
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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
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John and Peter are among the nine players a basketball coach
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23 May 2010, 15:38
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.
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Re: PS probability teams
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31 Jul 2008, 16:41
Total number of possibilities 9c5 Now say that John and Peter are already in the team. How can we pick rest of the three players out of 7 players  7C3
Hence 7c3/9c5 = 5/18



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Re: PS probability teams
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31 Jul 2008, 18:08
x97agarwal wrote: Total number of possibilities 9c5 Now say that John and Peter are already in the team. How can we pick rest of the three players out of 7 players  7C3
Hence 7c3/9c5 = 5/18 that's exactly what i did like 3 times, why am i not getting exactly 5/18? 7c3=35 9c5=126? oh i completely missed that 126 was divisible by 7...so it reduces to 5/18.



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Re: PS probability teams
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23 May 2010, 15:05
How can we solve this in probability approach. I thought the answer would be 1/9 x 1/8 x 3/7.



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Re: PS probability teams
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23 May 2010, 17:48
Thanks a lot.
Since the order doesn't matter isnt James A Jacob AA or A James Jacob A A, same? When do we need to use the multiplication factor (here 20) ingeneral? Also What does probability (1/9 * 1/8 * 3/7) represent



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Re: PS probability teams
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24 May 2010, 02:03
tochiru wrote: Thanks a lot.
Since the order doesn't matter isnt James A Jacob AA or A James Jacob A A, same? When do we need to use the multiplication factor (here 20) ingeneral? Also What does probability (1/9 * 1/8 * 3/7) represent Order does not matter YES, but favorable scenario JPAAA (JohnPeterAnyAnyAny) can occur in 5!/3! # of ways: JPAAA (J  first, P second, any for the third, ... ) the probability of this particular scenario would be: \(P=\frac{1}{9}*\frac{1}{8}*1*1*1=\frac{1}{72}\); JAPAA (J  first, any (but P) for the second, P  third, ...) the probability of this particular scenario would be: \(P=\frac{1}{9}*\frac{7}{8}*\frac{1}{7}*1*1=\frac{1}{72}\); ... ... ... There can be total of 20 such scenarios (20 as # of permuations of JPAAA is \(\frac{5!}{3!}=20\)) and each will have the probability of \(\frac{1}{72}\). The final probability would be the sum of all these favorable scenarios: \(20*\frac{1}{72}=\frac{5}{18}\). Hope it's clear. You can check the Probability and Combination chapters of the Math Book (link below) for more. Also check my posts at: probabilitycoloredballs55253.html#p6375254redchipsand2bluechips85987.html#p644603probabilityqsattention88945.html#p671958pc88431.html?highlight=probability+of+occurring+eventprobability88069.html?highlight=probability+of+occurring+eventcombinationproblemprincentenreview2009bin4q287673.html?highlight=probability+of+occurring+eventprobabilityproblem90124.html#p684229probability85523.html#p641151
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Re: PS probability teams
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24 May 2010, 14:14
Please explain why the permutation of jpaaa is not 5!.
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Re: PS probability teams
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24 May 2010, 14:30
Jinglander wrote: Please explain why the permutation of jpaaa is not 5!.
Posted from my mobile device Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is: \(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\). For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word. Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice. Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\). So according to the above # of permutations of 5 letters JPAAA would be 5!/3!, as there are 5 letters out of which A is represented thrice. Hope it's clear.
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Re: PS probability teams
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24 May 2010, 14:45
Still a bit confused don't all the a's represent distinct items as they are different ways to write jpaaa as is jpa1a2a3 is not the same as jpa3a2a1
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Re: PS probability teams
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24 May 2010, 15:19
Jinglander wrote: Still a bit confused don't all the a's represent distinct items as they are different ways to write jpaaa as is jpa1a2a3 is not the same as jpa3a2a1
Posted from my mobile device Not so. We are counting favorable scenarios to choose J and P. {J}{P}{A1}{A2}{A3} and {J}{P}{A3}{A2}{A1} represent one scenario: we choose J first, P second, any for third, any for fourth and any for fifth. Check the links in my previous post for several similar problems to practice. Also if you are not comfortable with probability approach you can always use combinatorics method. Luckily probability questions can be solved in many ways, choose the one which you are most comfortable with.
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Re: PS probability teams
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25 May 2010, 16:38
Ok I think I may have found a simplar way to solve this but not sure if it works given the events are nit independent. Ok John will make the team 5/9 of the time or 5/9 of all permutations of the team. Haven chosen the first person Peter will make the team 4/8 of the time. 5/9 * 4/8 yields the correct answer but I am not sure if I got the right answer by luck or this is a fair solution.



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Re: PS probability teams
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25 May 2010, 17:41
THANK YOU ALL FOR THIS..IT MAKES MUCH SENSE NOW



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Re: John and Peter are among the nine players a basketball coach
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21 Feb 2014, 08:50
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



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Re: John and Peter are among the nine players a basketball coach
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23 Feb 2014, 02:14
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 why not an easier solution? everyone in the group has a chance of 5/9 to be included in the team and the second person has the chance of 4/8. 5/9 x 4/8 = 5/18



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Re: PS probability teams
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10 Mar 2014, 10:49
Dear Bunuel
Can we do this by 1prob. of john and jack together. I tried it but the result is not coming out right somehow.



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Re: PS probability teams
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10 Mar 2014, 10:58



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Re: John and Peter are among the nine players a basketball coach
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10 Mar 2014, 20:29
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



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Re: John and Peter are among the nine players a basketball coach
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10 Mar 2014, 22:28
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
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Re: John and Peter are among the nine players a basketball coach
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10 Mar 2014, 23:28
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.




Re: John and Peter are among the nine players a basketball coach &nbs
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