There is an urn with three red balls, three blue balls, and three green balls. Suppose three balls are randomly drawn from the nine balls, and then three more balls will be randomly drawn from the six that were not selected on the first draw. What is the probability that the balls in the first subset drawn will all be the same color, and that the balls in the second subset drawn will all be the same color?

Attempted Solution:

So the first ball can be any color, but then the next two would have to be the same color as the first ball so we have,


Then the fourth ball can be any of the remaining six, but the next two balls will have to be the same color as the fourth ball so we have,

$\frac{9*2*1*6*2*1}{9*8*7*6*5*4}$ = $.003571$

Is this a valid solution?


Your solution seems correct to me. Alternatively, what we could do in the first question is to pick a color in any one of three ways. Once this is done, the three balls to be picked are fixed, so the answer is then $3 \times \frac{3 \times 2 \times 1}{9 \times 8 \times 7}$, the same as yours.

Similarly, in the second question, pick two colors, this is done by excluding one color, which happens in three ways. Now, fix which colour you choose first, this is done in two ways. Finally, this means all balls of the first color will come first, and then those of the second colour, so we have $3 \times 2 \times \frac{3 \times 2 \times 1 \times 3 \times 2 \times 1}{9 \times 8 \times 7 \times 6 \times 5 \times 4}$, the same answer as yours.

The difference in the solutions is that you pick a ball at random and see it's color, while I pick a color at random and fix the balls of that color.

  • $\begingroup$ Thanks for the confirmation. Always good to see alternative solutions as well. $\endgroup$ – Remy Sep 19 '17 at 3:34
  • $\begingroup$ You are welcome! $\endgroup$ – Teresa Lisbon Sep 19 '17 at 3:35

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