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### Zazuzu says

it depends on whether you replace the pens after taking them out.

the probability of obtaining 3 red pens if you replace the pen each time is:

(6/9)*(6/9)*(6/9)=8/27

the probability of obtaining 3 red pens if you don't replace each time is:

(6/9)*(5/8)*(4/7)=5/21

the probability of obtaining 2 red pen and 1 blue pen if you replace each time is:

(1/3)*[(2/3)*(2/3)]+(2/3)*[(1/3)*(2/3)+(2/3)*(1/3)]=4/9

the probability of obtaining 2 red pens and 1 blue pen if you don't replace each time is:

(3/9)*[(6/8)*(5/7)]+(6/9)*[(3/8)*(5/7)+(5/8)*(3/7)]=15/28

note: the trick to solving the last 2 problems was to consider every possible way of picking out 2 red pens and 1 blue. You could go red blue red ; red red blue ; blue red red; etc.

If your interested in that kind of thing, you can google combinatorics.