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PowerPrep Problem

by slimsohn » Fri Mar 09, 2007 12:30 pm
Hi, can someone help me with this problem- it seems simple, but I must be mistaken in my logic. Its from the Powerprep software Exam 2.

A certain basket contains 10 apples, 7 of which are red and 3 of which are green. If 3 different apples are to be selected at random from the basketbal, what is the probability that 2 of the apples selected will be red and 1 will be green?

A) 7/40
B) 7/20
C) 49/100
D) 21/40
E) 7/10

I thought it would just be 7/10 * 6/9 * 3/8, but thats wrong.
I'll post the answer after some discussion -- thanks.
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Re: PowerPrep Problem

by gabriel » Fri Mar 09, 2007 2:05 pm
slimsohn wrote:Hi, can someone help me with this problem- it seems simple, but I must be mistaken in my logic. Its from the Powerprep software Exam 2.

A certain basket contains 10 apples, 7 of which are red and 3 of which are green. If 3 different apples are to be selected at random from the basketbal, what is the probability that 2 of the apples selected will be red and 1 will be green?

A) 7/40
B) 7/20
C) 49/100
D) 21/40
E) 7/10

I thought it would just be 7/10 * 6/9 * 3/8, but thats wrong.
I'll post the answer after some discussion -- thanks.
... the answer shuld be 7c2*3c1/10c3.... which boils down to 21/40... so the answer is d.....

its 3:30 (thats am ) in india so not posting the detailed solution...if any one needs it ... will do it 2morrow.. tc
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by Badri » Fri Mar 09, 2007 4:13 pm
Assuming that the balls are not replaced in the basket after drawing:
There are three ways the incident can happen:


----1st Ball----2ndBall----3rd Ball---Probability
(1) Red--------Red--------Green-----P1 = ( 7/10 )*( 6/9 )*( 3/8 )
(2) Red--------Green------Red-------P2 = ( 7/10 )*( 3/9 )*( 6/8 )
(3) Green------Red--------Red-------P3 = ( 3/10 )*( 7/9 )*( 6/8 )


Total probability is P1 + P2 + P3
= 3*( 7*6*3 )/( 10*9*8 )
= 21/40
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by Cybermusings » Tue Mar 27, 2007 3:07 am
I think it's pretty simple...Let's C if I am right!

7C2 * 3C1 / 10C3

= 21/40

= Choice D

Cheers
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by mmslf75 » Sun Dec 20, 2009 9:21 am
Badri wrote:Assuming that the balls are not replaced in the basket after drawing:
There are three ways the incident can happen:


----1st Ball----2ndBall----3rd Ball---Probability
(1) Red--------Red--------Green-----P1 = ( 7/10 )*( 6/9 )*( 3/8 )
(2) Red--------Green------Red-------P2 = ( 7/10 )*( 3/9 )*( 6/8 )
(3) Green------Red--------Red-------P3 = ( 3/10 )*( 7/9 )*( 6/8 )


Total probability is P1 + P2 + P3
= 3*( 7*6*3 )/( 10*9*8 )
= 21/40
Should we consider,
6 aspects to this ? or just taht both REDs are same?
What am i missing here ?
R1 R2 G
R2 R1 G
R1 G R2
R2 G R1
G R1 R2
G R1 R2
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by papgust » Sun Dec 20, 2009 6:54 pm
I think we do not need to consider R1, R2 order here. Because a red ball is a red ball. You cannot distinguish 2 red balls.
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by mmslf75 » Tue Dec 22, 2009 9:41 am
papgust wrote:I think we do not need to consider R1, R2 order here. Because a red ball is a red ball. You cannot distinguish 2 red balls.
Aha !! May be that's coz we are not interested in "FIRST RED then GREEN" or vice versa..and that order is immaterial so long its RED colored

Wat sy ?
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by Testluv » Tue Dec 22, 2009 10:46 am
It is up to you whether or not you want to consider order. It doesn't matter, so long as you are consistent. gabriel and cybermusings treated the selections as simultaneous; badri considered order. Both approaches are correct. Personally, I like to treat these problems as simultaneous; pulling out one red ball after another without replacing them is the same thing as pulling out many red balls as a group at the same time, and you speed things up by not considering order.
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