a quality control engineer at a light bulb plant estimated that 1% of the bulbs it sells are defective. if you purcgase a package of 3 bulbs what is the probability that exactly one of them is defective?

To find the probability that exactly one of the three bulbs is defective, we can use the concept of binomial probability.

The binomial distribution is appropriate here because it involves a fixed number of trials (purchasing 3 bulbs) and each trial has two possible outcomes (defective or non-defective).

The formula for calculating the binomial probability of exactly one success (defective bulb) in n trials is:
P(X = k) = nCk * p^k * (1-p)^(n-k)

Where:
P(X = k) represents the probability of getting exactly k successes.
nCk represents the number of combinations, also known as the binomial coefficient, calculated as n! / (k!(n-k)!).
p represents the probability of success in a single trial (1% or 0.01).
k represents the number of successes (one defective bulb).
n represents the number of trials (3 bulbs).

Let's plug in the values and calculate the probability:

P(X = 1) = 3C1 * (0.01)^1 * (1-0.01)^(3-1)
P(X = 1) = 3 * (0.01) * (0.99)^2
P(X = 1) = 0.03 * 0.9801
P(X = 1) ≈ 0.0294

Therefore, the probability that exactly one of the three bulbs is defective is approximately 0.0294 or 2.94%.