Binomial distribution question about defective light bulbs

parabola

New member
A company producing light bulbs knows that the probability that a new light bulb is defective is 0.5%.

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!

tkhunny

Moderator
Staff member
Did you solve the problem manually, just to prove that your calculator button-pushing didn't skip anything?

You could also build a quick spreadsheet to specify the entire distribution and prove your answer to be correct.

Part of the learning of mathematics is the confidence required to refute bad results. Also, one needs to work so that results can be followed an verified.

parabola

New member
Hi,
Thank you for your reply! I tried solving it manually as well and got the same result as with the calculator function. I should have said that in my original message, sorry I'm confident that I'm calculating the probability correctly but could you confirm that I'm using the right distribution to begin with, X~B(20, 0.0296)? Thank you so much.