[Math] the probability of drawing 2 same colored marbles

probability

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

Best Answer

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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