Question

There are 6 red balls and 4 green balls in a bag. If 3 are picked out at random, what is the probability that you pick 1 red and then 2 greens if the balls are not replaced?

Answer 1

We have 6 red balls and 4 green balls. We want to know the probability of picking 1 red ball and 2 green balls without replacements in any order of result.

There are 3 possible orders of results that we're interested (rrg, rgr and grr)

We just want to calculate the probability of each case.

P(rrg) = (6/10)*(5/9)*(4/8) = 1/6 or 16.66..%

P(rgr) = (6/10)*(4/9)*(5/8) = 1/6 or 16.66..%

P(grr) = (4/10)*(6/9)*(5/8) = 1/6 or 16.66..%

Therefore: P(2r+1g) = P(rrg) + P(rgr) + P(grr) = 3/6 = 1/2 or 50%

Notice that also P(2r + 1g) = 3*P(rrg), since order doesn't matter.