Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 18 Jun 2010
Posts: 300
Schools: Chicago Booth Class of 2013

A basketball coach will select the members of a fiveplayer [#permalink]
Show Tags
17 Aug 2010, 11:27
1
This post was BOOKMARKED
Question Stats:
71% (02:08) correct
29% (01:26) wrong based on 58 sessions
HideShow timer Statistics
A basketball coach will select the members of a fiveplayer 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 OPEN DISCUSSION OF THIS TOPIC IS HERE: abasketballcoachwillselectthemembersofafiveplayer132143.html
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 12 May 2014, 00:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



Senior Manager
Status: Fighting on
Joined: 14 Mar 2010
Posts: 316
Schools: UCLA (R1 interviewWL), UNC(R2interviewding) Oxford(R2Admit), Kelley (R2 Admit $$), McCombs(R2)
WE 1: SE  1
WE 2: Engineer  3

Re: MGMAT Probability [#permalink]
Show Tags
17 Aug 2010, 14:11
you can but that will require more cases and will be time consuming.
straight forward way is to group J and P as always present on the team and as order does not matter so we just need to find the total number of ways to select rest three players = 7c3
total number of ways of selecting 5 players out of 9 = 9c5
probability = 7c3/9c5 = 5/18



Senior Manager
Joined: 18 Jun 2010
Posts: 300
Schools: Chicago Booth Class of 2013

Re: MGMAT Probability [#permalink]
Show Tags
17 Aug 2010, 23:22
ingoditrust wrote: you can but that will require more cases and will be time consuming.
probability = 7c3/9c5 = 5/18 Look at my approach  it's easy, nice and not at all time consuming. But I'm somehow getting the wrong answer I decided to find the probability of not getting these two guys in the choosen group and then substruct it from 1. What's wrong with my approach? 1(7/9)*(6/8)*(5/7)*(4/6)*(3/5)= 15/18



Intern
Joined: 15 Aug 2010
Posts: 23
Location: Mumbai
Schools: Class of 2008, IIM Ahmedabad

Re: MGMAT Probability [#permalink]
Show Tags
18 Aug 2010, 00:30
Financier wrote: I decided to find the probability of not getting these two guys in the choosen group and then substruct it from 1. What's wrong with my approach?
Prob of choosing both J and P = 1  ( Prob of choosing neither J nor P + Prob of choosing J and not P + Prob of choosing P and not J) Hope this helps. This was highlighted in one of the other probability threads earlier this day only. Thanks.
_________________
Naveenan Ramachandran 4GMAT  Mumbai



Senior Manager
Joined: 15 Aug 2013
Posts: 311

Re: MGMAT Probability [#permalink]
Show Tags
11 May 2014, 17:30
I'm not sure if I'm misunderstanding the question but the way I read it, the question states that "what is the probability that jon and peter" will be selected. Correct?
If that's the case, why isn't this method working:
(5c2)/(5c9)  I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?
Additionally, I tried the direct probability approach and that seemed to work out:
(2/9)(1/8)(5!/2!3!)  This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...
This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this?



Math Expert
Joined: 02 Sep 2009
Posts: 39751

Re: MGMAT Probability [#permalink]
Show Tags
12 May 2014, 02:25
russ9 wrote: I'm not sure if I'm misunderstanding the question but the way I read it, the question states that "what is the probability that jon and peter" will be selected. Correct?
If that's the case, why isn't this method working:
(5c2)/(5c9)  I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?
Additionally, I tried the direct probability approach and that seemed to work out:
(2/9)(1/8)(5!/2!3!)  This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...
This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this? About 5C2 in the numerator: are we choosing 2 from 5? No, we want to choose 5 players, including John and Peter, from 9 players. About the probability approach: though the order of the selection does not matter, John and Peter can be any from the 5 selected. We can have {John, Peter, any, any, any}, {John, any, Peter, any, any}, {John, any, any, Peter, any}, ... Now, each of these cases have equal probabilities to occur: 1/9*1/8*1*1*1 and there can be 5!/3! cases, so the overall probability is 1/9*1/8*1*1*1*5!/3! = 5/18. A basketball coach will select the members of a fiveplayer 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 Total # of fiveplayer teams possible is \(C^5_9=126\); # of teams that include both John and Peter is \(C^2_2*C^3_7=35\); \(P=\frac{35}{126}=\frac{5}{18}\). Answer: D. OPEN DISCUSSION OF THIS TOPIC IS HERE: abasketballcoachwillselectthemembersofafiveplayer132143.html
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Re: MGMAT Probability
[#permalink]
12 May 2014, 02:25







