There are 10 balls in a bag, 6 red and 4 green. If three are picked out at random what is the probability of one red and two green?

Question

Based on probability

Answer 1

First, we need to count all of the ways that 3 balls (without replacement, regardless of color) can be drawn from a set of 10. This is simply 10 choose 3.

 

Second, we need to count the ways exactly one red and two green can be chosen. This is (6 choose 1)*(4 choose 2).

 

So our probability is: (6 choose 1)*(4 choose 2) / (10 choose 3).

 

I like to simulate the experiment in R 1000 times to see if my math works out:

 

> choose(6, 1)* choose(4, 2) / choose(10, 3)
[1] 0.3
>
> bag <- rep(c("red","green"), each = 6, length=10)
> red1.green2 <- replicate(1000, sum(sample(bag, 3) == "red") == 1)
> mean(red1.green2)
[1] 0.295

 

So we see that we got the desired event in 295 of 1000 experiments, so the probability 0.3 we calculated is likely to be correct.