need help solving it

Question

An urn contains 10 red, 11 white and 20 blue marbles. A child selects two marbles at random and without replacement from the urn.

Find the probability that the colors of the selected marbles are different.

Answer 1

Hi Lakotah,

This is a combination of the Addition Rule and Multiplication Rule, but first, we need to establish how we get two different colors. There are three possibilities: red and white, red and blue, and white and blue. Abbreviated, this means the options are:

RW, WR, RB, BR, WB, BW

Now, there are a total of 41 marbles in the box, so we need to compute the probabilities for each scenario above and add the result together. I will do the first two and let you handle the rest:

P(RW) = (10/41)*(11/40) = 0.067

Remember that we removed one marble, so we subtract one from the total to get the second probability.

P(WR) = (11/41)*(10/40) = 0.067

Repeat this process for RB, BR, WB, and BW. Then, add all of those probabilities together to get your final answer. I hope this helps.