A bag contains 3 red marbles, 6 blue marbles and 8 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 1000th, that all three marbles drawn will be green?

Answer 1

OK, this question has two possible solutions. What if the marble drawer puts the marble back after each draw? This is called "replacement" and this will give a different answer from "no replacement".

So let's first consider this problem WITH replacement:

With 17 total marbles, and 8 of them are green, the probability of picking a green marble once is simply 8/17. But then if you repeat this, then you must multiply this probability for each draw. This gives

(8/17) * (8/17) * (8/17), or you could write (8/17)ˆ3 (cubed)

A way to remember that you must multiply here, rather than add, is that if you were to add each time, you would get a probability greater than 1, and that can't happen. Probabilities are always between 0 and 1.

OK, now WITHOUT replacement:

The first draw is the same as before: the probability of picking a green marble is 8/17.

But the second draw is different. Now there are only 16 marbles left, and only 7 of them are green, so the probability that the second pick is green is 7/16.

Likewise, the probability of the third draw being green is 6/15.

So as before, all of these probabilities must be multiplied to solve for the "union", or the probability of all 3 events happening in this way. So the probability of picking 3 green marbles without replacement is calculated as:

(8/17) * (7/16) * (6/15)