00:01
This problem says a box contains two red crayons, six blue, five green, nine yellow, and three purple.
00:06
And we're told a crayon is drawn at random, not replaced, and then another is drawn.
00:11
And we're given two probabilities to find.
00:13
We want to find the probability that we select a yellow, and then a green, and also the probability that we select blue for both.
00:19
And again, both of these probabilities are set up to not replace the crayon after the first choice.
00:23
So for our first probability, we're just going to focus on the probability that we get a yellow crayon, and there's nine yellow total out of the total when we add up our total numbers, or total colors of 25.
00:35
So we have a 9 in 25 chance of getting yellow for the first crayon.
00:40
And then we need to remember that we don't put this crayon back.
00:43
So when we look at the probability now that we have a green, we still have five green in the box, but one of our total has been taken away, because we assume that we had a success for pulling a yellow from the 25 to start.
00:56
And with that yellow gone, it doesn't affect our number of green, but it does affect the total in the box, which would be over 24.
01:03
And once we can, or once we have these two fractions set up, we can simplify a little bit by showing 5, cancelling 25 to be 1 over 5, when you divide them both by 5.
01:14
And then 9 and 24 are both divisible by 3.
01:16
9 divided by 3 is 3...