in a bag of marbles, there are 12 green, 12 blue, and one red. Find the probability for each situation as a decimal rounded to the nearest hundredth.

Question

What are the chances you pull a blue or green on your first and second try without replacing them, then pull the red?

What are the chances of pulling the red on either your first, second, or third try without replacing the marbles in-between?

Answer 1

As there are 24 blue and green and one red, P(blue or green then red) = P(24,2)*1/(P(25,3) = 24*23/(25*24*23) = 1/25 = .04

There are two ways of getting the chance of a red on any selection. This also gives us a different way to solve the first problem.

3 * .04 = .12

Alternatively, you are selecting 3 from 25 without replacement, so the chance of the one red being in the 3 has to be 3/25

This also suggests another way to do the first problem. Rather than figuring first the probability of getting a blue or green, then the red, just figure the probability that the third selection is a red. This is 1/25. Then, since there is only 1 red, you are guaranteed that the other two will not be red, so the probability of two non-red then red is 1/25.