00:01
For this problem, to begin, i'll write down our events as b1 and b2, where b1 is the event that the first ball is blue, and b2 is similarly the event that the second ball is blue.
00:19
Now, since we know that the balls are being randomly sampled without replacement, for part a, the probability of b1 and b2 is going to be basically found by taking into account the idea that the probability of b2 is going to be conditional on the outcome of that first draw.
00:39
So, probability of b1 and b2 is going to be equal to the probability of getting a blue ball on the first draw times the probability of getting a blue ball on the second draw, given that we got a blue on the first.
00:54
Now, probability of b1 is going to be the number of blue balls in the urn overall, which we start out with rather, which is 5, divided by the total number of balls in the urn, 5 plus 6 plus 7, for a result of 18.
01:11
Then, for probability of b2 given b1, well, we know that we've removed one of the blue balls from the urn, so there are four blue balls remaining, and 17 overall...