Final answer:
The probability that exactly 3 of 8 light bulbs are defective when the probability that a single bulb is defective is 0.1 is found using the binomial probability formula and is approximately 3.11%.
Step-by-step explanation:
The question asks to find the probability that exactly 3 of 8 light bulbs are defective when the probability that a single light bulb is defective is 0.1. This is a binomial probability problem because there are a fixed number of trials (8 light bulbs), each trial can result in just two possible outcomes (defective or not defective), the probability of a defective light bulb is the same for each trial, and the trials are independent.
To find the probability of exactly 3 defective out of 8, we use the binomial probability formula:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n is the number of trials (8)
- k is the exact number of successes (3 defective bulbs)
- p is the probability of success (0.1)
Using the formula, we calculate:
P(X=3) = (8 choose 3) * 0.1^3 * 0.9^5
First, calculate (8 choose 3) which is the number of ways to choose 3 defective bulbs from 8, which is 56. Then multiply this by 0.1^3 (the probability of having 3 defective bulbs) and by 0.9^5 (the probability of the remaining 5 bulbs being non-defective). So, P(X=3) = 56 * 0.001 * 0.59049 = 0.031104, or about 3.11%.