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24 May 2006, 01:07
Kudos
Bookmarks
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
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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24 May 2006, 01:31
Kudos
Bookmarks
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
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
24 May 2006, 02:27
Kudos
Bookmarks
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
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
