Suppose there are 3 colored marbles (Red, Blue, Green). There are 10 red, 20 blue, 30 green marbles. You select 2 marbles without replacement. What is the probability that they are the same colors?

My solution:
$10 \choose 2$ ways to choose 2 red
$20 \choose 2$ ways to choose 2 blue
$30 \choose 2$ ways to choose 2 green $60 \choose 2$ ways to choose either red, blue, or green marbles without replacement

So the probability is $\frac{{10 \choose 2}+{20 \choose 2}+{30 \choose 2}}{60 \choose 2} $

I'm not sure if it makes sense to add the combinations in the numerator

  • 2
    $\begingroup$ Your solution is okay. The addition makes sense because it concerns the probability of a union of disjoint events. $\endgroup$ – drhab Mar 5 '15 at 19:10

There are two ways to think of this. The first is that if we define events $R$ to be two red balls, $B$ to be two blue balls, and $G$ to be two green balls, then we can verify that:

  • $P(R) = \big(\frac{10}{60}\big)\big(\frac{9}{59}\big) = \frac{3}{118} \approx 0.025$
  • $P(B) = \big(\frac{20}{60}\big)\big(\frac{19}{59}\big)= \frac{19}{177} \approx 0.107$
  • $P(G) = \big(\frac{30}{60}\big)\big(\frac{29}{59}\big) = \frac{29}{118} \approx 0.246$

Then, $P(R \cup B \cup G) = P(R) + P(B) + P(G) = \frac{67}{177} \approx 0.379$

Note that each of the events $R,, B, G$ are disjoint and thus we may use the additive property.

Second, if you would like to do this with counting then you simply compute $P(R)$ (and for the others) in terms of counting ratios as you did and note that they sum to precisely what you computed. You can cross check with what I presented here and you will find they are the same.

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$10 \over 60$,$20 \over 60$, and $30 \over 60$ are the respective probabilities of drawing each of the respective colors from the bag, for the first draw. Since you want the probability that you get the same color after the first draw, there will be one less marble in each of the respective sets and one less marble in the total number of marbles. This is because you took a marble out and didn't replace it. Those probabilities are $9 \over 59$,$19 \over 59$, and $29 \over 59$. The probability is given by probability of A & B which equals A*B of the two draws, and by probability of A or B or C which equals A+B+C (Read up on Mutually Exclusive and Dependent events). This is given by, $${10 \over 60}*{9 \over 59}+{20 \over 60}*{19 \over 59}+{30 \over 60}*{29 \over 59}$$ Which gives a probability of about 38%. Using combinations really isn't the best way in this case because this isn't a how many possibilities are there problem.

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