a box contains six red pens and three blues.find the probability of obtaining three red pens and...

two red pens and one blueit 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.

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