Question

Please help with this?
Three light bulbs are chosen at random from 15 bulbs of which 5 are defective. Find the probability that exactly one is defective.
Thanks!

Answer 1

exactly one defective means one is and two are not
the number of ways you can pick one out of 5 defective is 5
the number of ways you can pick two out of 10 not defective is \(\binom{10}{2}=\frac{10\times 9}{2}=5\times 9=45\)
so your numerator is \(5\times 45\) while your denonimator is the number of ways you can choose three out of 15 which is \(\binom{15}{3}\)

Answer 2

Thanks for the help! Greatly appreciate it. :)