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probability or combination

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Manager
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probability or combination [#permalink] New post 24 May 2006, 00:07
hello
plz need help on the following
when dealing with question like this one
source advanced math forum

The only contents of a bag are 4 pens that write blue and 3 pens that write green. If 4 of those pens are chosen at random, what is the probability that 2 of the pens write blue and 2 of the pens write green?

A) 2/7

B) 1/2

C) 18/35

D) 4/7

E) 9/14
I don 't know when to use simple probability or combination
usually permutations and combination deal with the number of ways

i am quite lost

thanks
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Senior Manager
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 [#permalink] New post 24 May 2006, 00:31
Probability of certain outcome is always equal to the number of
successful outcomes divided by the number of total possible outcomes.

Total outcomes = 7C4
Successful = 4C2 (for blue pens ) * 3C2 ( for green ones )

4C2*3C2 / 7C4 = 18/35
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 [#permalink] New post 24 May 2006, 01:27
Method One: Using Simple probability

4 pens in which 2 will write blue and 2 will write green. Hence possible combinations

BBGG
BGBG
GBGB
GGBB
GBBG
BGGB

Each has a probability of 3/35 (from 4/7*3/6*3/5*2/4)

Hence total probability = 6*3/35 = 18/35

Method 2:

THere are 7C4 ways 4 pens can be chosen.
two blue can be chosen 4C2 ways and 2 green can be chosen in 3C2 ways

Therefore, P = (4C2*3C2)/7C4 = 18/35
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 [#permalink] New post 24 May 2006, 01:38
Agree with 18/35.

Even I had problems with these type of questions as one gets confused whether to treat it as 'with replacement' or without.

I think the approach is to treat each similar coloured object (pens, marbles etc) as unique - calculate number of combinations for one colour - then mutiply with number of combinations for other colour to get total number of ways for the condition to be true.

To get probability we devide with total number of ways possible, which is given by nCr.
  [#permalink] 24 May 2006, 01:38
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