confusion

**bluffmaster.roy.007** · August 15th 2009, 01:46 AM

In a bottle of 3 red, 4 green and 5 blue marbles, if 2 marbles are taken out, what is the probability that two marbles are of red color

my answer is 1/4 i just did like number of red balls by total number balls .

i have another confusion like 3c2/12c2 which is 1/22...which is correct

**mr fantastic** · August 15th 2009, 02:14 AM

Originally Posted by bluffmaster.roy.007

In a bottle of 3 red, 4 green and 5 blue marbles, if 2 marbles are taken out, what is the probability that two marbles are of red color

my answer is 1/4 i just did like number of red balls by total number balls .

i have another confusion like 3c2/12c2 which is 1/22...which is correct

I assume the balls are taken out without replacement. In which case a tree diagram easily eveals the answer: $\frac{3}{12} \cdot \frac{2}{11}$ which can be simplified.

**e^(i*pi)** · August 15th 2009, 02:17 AM

Originally Posted by bluffmaster.roy.007

In a bottle of 3 red, 4 green and 5 blue marbles, if 2 marbles are taken out, what is the probability that two marbles are of red color

my answer is 1/4 i just did like number of red balls by total number balls .

i have another confusion like 3c2/12c2 which is 1/22...which is correct

There are a total of $3+4+5 = 12$ marbles in the bag, 3 of these are red.

Taking out one marble means there is a $\frac{3}{12} = \frac{1}{4}$ chance that it is red. Also there are now only 11 marbles in the bag

The chance that this is also red is $\frac{2}{11}$ because there are only 2 red marbles as the previous one was red.

Since these are an "and" example you multiply these together to give the chance of two red ones: $P(RR) = \frac{1}{4} \times \frac{2}{11}$

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