Please explain how she got 0.6374. She said solve it in your own, but I am not too sure how she arrived to that solution.

Example 4 Suppose a box contains 3 defective light bulbs and 12 good bulbs. Suppose we draw a simple random sample of 4 light bulbs, 1. What is the probability that none of bulbs drawn are defective? 0 3 bad 12 good 15 4 4 2. What is the probability that at least one of the bulbs drawn is defective? Plat least I is defective) = P(1 Bad) + P(2 Bad) + P(3 Bad) + P(4 Bad) = 1- P(0 Bad) 0.6374 UNIVERSITY of HOUSTON DEPARTMENT OF MATHEMATICS