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baseball team probability

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baseball team probability

by canuckclint » Sat Oct 25, 2008 9:16 pm
John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

OA to come soon!
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easier question

by canuckclint » Sat Oct 25, 2008 9:21 pm
An easier question to go along for the keeners.

A baseball card decreased in value 20% in its first year and 10% in its second year. What was the total percent decrease of the card's value over the two years?
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Re: baseball team probability

by sudhir3127 » Sat Oct 25, 2008 9:40 pm
canuckclint wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

OA to come soon!
its 5/18 for me ...

Total number of ways one can choose 5 people from 9 = 9c5

If John and Peter are already there, then we need to select 3 from remaining 7 = 7C3.

hence answer = (7C3)/(9C5)= 5/18..

hope it makes sense.. do let me know if u have any doubts..
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by canuckclint » Sat Oct 25, 2008 10:12 pm
If John and Peter are already there, then we need to select 3 from remaining 7 = 7C3.

I don't get the above statement.
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by bhanu75 » Sun Oct 26, 2008 3:07 am
The statement of sudhir3127 means:

As John and Peter have already been selected, so you need to chose only 3 other players out of remaining total of 7 (9 - John - Peter), which is 7C3.
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by canuckclint » Sun Oct 26, 2008 8:09 am
Yes but the thing is they are not chosen by looking at them.
They are chosen by random.
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by scoobydooby » Sun Oct 26, 2008 9:28 am
theres only one way in which Peter can be chosen (only one person called Peter) and only one way in which John can be chosen (only one person called John)
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by stop@800 » Sun Oct 26, 2008 2:23 pm
Probability its 5/18

and for another one
total % decrease will be 28%

initailly 100
after 1st yr 80
after iind yr 80 - 8 = 72
total decrease = 100-72 = 28
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by bhumika.k.shah » Sun Mar 28, 2010 12:59 pm
sudhir3127 wrote:

hence answer = (7C3)/(9C5)= 5/18..
Why is it 7C3 / 9C5 ?
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by Stuart@KaplanGMAT » Sun Mar 28, 2010 4:17 pm
canuckclint wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

OA to come soon!
Please post answer choices and the source - thanks!
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by bhumika.k.shah » Sun Mar 28, 2010 7:53 pm
A.1/9
B.1/6
C.2/9
D.5/18
E.1/3

OA D

For me, the Source is [spoiler]MGMAT CAT # 2[/spoiler]
Stuart Kovinsky wrote:
canuckclint wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

OA to come soon!
Please post answer choices and the source - thanks!
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by thephoenix » Mon Mar 29, 2010 6:22 am
bhumika.k.shah wrote:
sudhir3127 wrote:

hence answer = (7C3)/(9C5)= 5/18..
Why is it 7C3 / 9C5 ?
7c3 becoz peter and jhon can be selected in only one way and rest 3 has to be selected from 7 available players(favourable outcomes)

9C5 because out of 9 , 5 has to be selceted (total possible selection or outcomes)

P=favourable outcome/tot outcome
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by eaakbari » Mon Mar 29, 2010 1:16 pm
There is 1 john prob of selecting him = 1c1
There is 1 peter; prob of selecting him = 1c1
There are 7 other players = 7c3

Total options = 9c5


Hence 1c1*1c1*7c3/9c5

equals 5/18
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by eaakbari » Mon Mar 29, 2010 1:19 pm
Second question
Value after 1st decrease = 0.8x
Value after 2nd decrease 0.9 *0.8x = 0.72x

Therefore total decrease = x-0.72x

That is 28%
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by YARAB800 » Fri May 07, 2010 7:32 am
This might be a stupid question, but what signaled you to think that this is a combination question? was it because the question said "five-player teams among 9"? The reason I got it wrong was because I saw the "probability" and was fixed on getting the answer using the concept of probabilities. Is there a way to answer this such questions using the rule of probability?

Thanks
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