A. Find the probability that a pack of 6 light bulbs contains at least one defective one.

B. Mario buys 20 packs of six light bulbs. Find the probability that more than 4 of the boxes contain at least one defective light bulb.

So I'm supposed to use binomial distribution and for part A I did X~B(6, 0.005) and P(X is greater or equal to 1) = P(X=0) = 0.0296 which is correct according to the answers.

For part B I've used binomial distribution as well so that X~B(20, 0.0296) but used the exact value for the probability. Then I used the binomcdf function on my TI-84 plus calculator to find P(X is greater than 4) = 1 - P(X is greater or equal to 4) and got 2.44 * 10

^{-4}. The answer is 5.02 * 10

^{-4}according to the book.

Then again this book has a lot of wrong answers in it but since my answer to part A is the same as the book's I think I am doing something wrong. I don't really know what that something is so any help would be appreciated thank you!