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Select 3 marbles at random from a jar with 6-blue, 4-green, and 3-red. What is probability that A.All are green? B.What is probability of selecting 2-blue and 1-red? C.What is probability of selecting in exact order 1 blue, 1 green, and 1 red?

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1 Answer 1

Presumably we are sampling without replacement.

A: There are $13$ balls, so $\dbinom{13}{3}$ ways to choose $3$ balls. All these ways are equally likely. There are $\dbinom{4}{3}$ ways to choose $3$ green. Divide.

B: Same denominator. There are $\dbinom{6}{2}$ ways to choose $2$ blue. For each of these ways, there are $\dbinom{3}{1}$ ways to choose $1$ red, for a total of $\dbinom{6}{2}\dbinom{3}{1}$. Now divide.

C: The probability that the first is blue is $\dfrac{6}{13}$. Given this has happened, the probability the next is green is $\dfrac{4}{12}$. Given these two things have happened, the probability the last is red is $\dfrac{3}{11}$. Multiply.

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Nicolas. Do you mind working it thru so i can compare to make sure i divided or multiplied correctly for each item? I appreciate you taking time to help me –  kimj Sep 29 '12 at 4:10
    
For first, reduced fraction $2/143$. For second, reduced fraction $45/286$. For third, your calculator is as good as mine. But maybe not. Mine cost about $10$ dollars (sale) $15$ years ago, still fine. –  André Nicolas Sep 29 '12 at 4:14
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