00:01
We're told in this question that a rare form of cancer occurs according to a poisson distribution with a mean of 0 .00011 per child.
00:13
For part a, we are asked to find the probability that the number, or actually, we are asked to find the mean number of cases out of 12 ,429 children and is equal to 12 ,429.
00:34
So the mean number of cases is equal to n times p, which is 12 -429 times .0011.
00:52
And this gives us .1367 approximately.
01:00
And for b we are asked to find the probability that the number of cases of this cancer in a group of 12 ,429 is 0 or 1.
01:18
This is equal to the probability that x is at most 1.
01:32
Now the probability mass function for the poisson random variable is given by this formula.
01:50
So continuing with this calculation.
01:54
For the first term x equals 0, e to the minus 0 .1367.
02:03
And for the second term where x is equal to 1, get this, which is approximately 0 .9915.
02:22
And then for c we're asked for the probability of more than one case...