00:01
For this problem, to begin, i'll note that for the situation described, we can consider the number of defective items as a binomial random variable where the number of trials, n, is equal to 20, and p, probability of a defect, is 0 .05.
00:23
For part a, we're looking for the probability that x equals 2.
00:27
So we can find this using the pmf, the probability mass function, or pdf, probability distribution function for a binomial.
00:36
It's going to be n choose the value of x, so 20 choose 2, times 0 .05 to the power of 2, times 1 minus p, so that's 0 .95, to the power of n minus 2, so to the power of 18.
00:54
Should give us a result of roughly 0 .19.
00:59
Then for part b, the probability that it is at least 3 items, so that's probability x greater than or equal to 3, we can find by taking 1 minus the probability that x is less than or equal to 2...