00:01
We have a sample of 100 and a probability of 1%.
00:04
So we're looking at a binomial distribution, so that'll look like the bin of 100 comma .01 for our distribution type.
00:12
And then b was the probability that at least one packet must be resent.
00:16
So that's going to be the probability of x is greater than equal to 1, which is equal to 1 minus the probability of exactly 0 that are sent back, which is going to be equal to 1, probably that zero sent back, so that's 0 .99 .99 % chance that we don't send any back, or that we don't send one back times probably of 100.
00:40
And that is going to be 63%.
00:49
And then for part c, we're looking at the ones that are greater than equal to 2.
00:54
And that is going to be one, and it's probably that we have zero, and probably that we have one.
01:00
And those respective probabilities are going to be that one we found before, this one there.
01:04
And then the one that we get wrong and then the 99 that we get right and that's going to be 0 .264.
01:12
Next step is to calculate the average, the mean, so that's going to be the expectation of x which is equal to n times p.
01:21
So that's going to be 100 times 1%, or 1.
01:27
And then our variance is going to be equal to n p times 1 minus p...