00:03
We are given the distribution of colors of peanut m &ms or 12 % of brown, 15 % yellow, 12 red, 23 blue, 23 orange, and 15 % green.
00:17
And we're going to compute the probability that the first orange candy is the third m &m selected.
00:25
So this is a geometric distribution.
00:29
Even if we don't know that, we can kind of reason through this.
00:31
So if the first orange candy is the third one, that means the first one would be, the first candy would not be orange.
00:43
That's one minus .23.
00:46
And that's twice.
00:48
And then we get our .23.
00:51
And we do that multiplication.
00:53
And we get this.
00:54
0 .136.
00:57
Now we'll find the probability that the first orange candy is a third or the fourth.
01:03
So we know the third is this.
01:07
The fourth would be 1 minus 0 .23 to the third times .23.
01:18
Put parentheses here just to keep those separate.
01:22
And so to get this or that and there's no overlap, we can add those two together.
01:27
So we get .241.
01:31
Now we're going to compete the probability that the first orange candy is among the first three.
01:34
So somewhere in the first three.
01:36
So it's going to be, i'm going to take this, so it's .23 times one minus .2, 3.
01:46
To the second...