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Manhattan GMAT probability / Combinatorics

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Manhattan GMAT probability / Combinatorics

by pkw209 » Tue Mar 02, 2010 3:34 pm
A brief explanation would be great. Thanks!

8 students have been chosen to play for the basketball team. If every person on the team has an equal chance of starting, what is the probability that both Tom and Alex will start? (Assume 5 starting positions)

answer is [spoiler]5/14[/spoiler]
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by stufigol » Tue Mar 02, 2010 9:48 pm
Number of total possibilities: Use Combinations.
Out of 8 players, we choose 5

8 choose 5
-> 8 C 5
--> 8! / (4! * 5!)

Let's move on to the number of possibilities we want.

Number of possibilities we want: All possible teams that include John and Peter.
A team has 5 people, and 2 of them are John and Peter. Therefore, we need to find all possible combinations for the other 3 empty spots. Since John and Peter are already chosen, we have 8-2=6 possible players that can fill those 3 spots. Use Combinations again.

Out of 6 players, we choose 3

6choose 3
-> 6C 3
--> 6! / (3! * 3!)

and u ll find the solution
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by yeahdisk » Wed Mar 03, 2010 1:36 am
stufigol wrote:Number of total possibilities: Use Combinations.
Out of 8 players, we choose 5

8 choose 5
-> 8 C 5
--> 8! / (4! * 5!)
Isn't 8 C 5

->8! / (5! * (8-5)!)

-->8! / (5! * 3!)

= 56

?
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by sanju09 » Wed Mar 03, 2010 1:45 am
pkw209 wrote:A brief explanation would be great. Thanks!

8 students have been chosen to play for the basketball team. If every person on the team has an equal chance of starting, what is the probability that both Tom and Alex will start? (Assume 5 starting positions)

answer is [spoiler]5/14[/spoiler]
When both Tom and Alex are certain to start, we are left to select 3 more students from the remaining 6 to join Tom and Alex for the start. This can be done in 6C3 ways, and the total ways of selecting all 5 starters is 8C5. Hence, the required probability

= 6C3/8C5

= [spoiler]5/14[/spoiler].
The mind is everything. What you think you become. -Lord Buddha



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by harsh.champ » Wed Mar 03, 2010 4:22 am
yeahdisk wrote:
stufigol wrote:Number of total possibilities: Use Combinations.
Out of 8 players, we choose 5

8 choose 5
-> 8 C 5
--> 8! / (4! * 5!)
Isn't 8 C 5

->8! / (5! * (8-5)!)

-->8! / (5! * 3!)

= 56

?

I agree with yeahdisk.

You could have got the wrong answer,stufigol.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by stufigol » Wed Mar 03, 2010 11:39 am
yeah my bad , thxs for correcting
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by pkw209 » Thu Mar 04, 2010 2:03 pm
Thanks. You guys provided the manhattan gmat explanation.

What are the other ways of solving this problem?
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by tata » Fri Mar 05, 2010 8:24 am
Experts,

Should we not consider Alex and Tom also be selected in 2 ways?
So shouldnt the # of possible outcomes be 6C3 * 2 instead of 6C3 only?
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by djkvakin » Fri Mar 05, 2010 10:48 am
The chance that Alex will be selected to start out of 8 players is 5*(1/8)=5/8. The chance that Tom would be selected out of the 7 players left is 4*(1/7)=4/7.

The chance that both would be selected is 5/8*4/7=20/56=5/14
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by sanju09 » Sat Mar 06, 2010 12:40 am
tata wrote:Experts,

Should we not consider Alex and Tom also be selected in 2 ways?
So shouldnt the # of possible outcomes be 6C3 * 2 instead of 6C3 only?
When we select, we seldom permute.
The mind is everything. What you think you become. -Lord Buddha



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by pkw209 » Wed Mar 31, 2010 2:34 pm
I got the answer this way:

5 spots, Alex is one of the spots so 5C1 = 5

4 remaining spots, Tom is one of them so 4C1 = 4

5 x 4 = 20

total possibilities ----> 8C5 = 56

20/56 = 5/14
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by eaakbari » Thu Apr 01, 2010 1:37 am
(1C1 * 1C1 * 6C3) / 8C5

= 5/14
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