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 bulb in a package of three is defective, we can use the concept of binomial probability. The binomial distribution calculates the probability of a fixed number of successes in a series of independent trials with the same probability of success.
In this case, the probability of a bulb being defective is given as 1% or 0.01. Let's denote this probability as P(defective) = 0.01. The probability of a bulb being non-defective is therefore 1 - P(defective) = 1 - 0.01 = 0.99.
Now, we need to apply the binomial distribution formula:
P(X = k) = (n C k) * P^k * (1 - P)^(n - k)
Where:
- P(X = k) is the probability of exactly k defects
- "n C k" is the number of combinations of taking k objects from a set of n objects, given by n! / (k! * (n - k)!)
- P is the probability of success (defective bulb), which is 0.01
- n is the number of trials (bulbs in the package), which is 3
- k is the number of successes (defective bulbs), which is 1
Applying the values into the formula, we have:
P(X = 1) = (3 C 1) * 0.01^1 * 0.99^(3 - 1)
= (3 * 2 / (1 * 2)) * 0.01 * 0.9801
= 3 * 0.01 * 0.9801
= 0.0294
Therefore, the probability that exactly one bulb is defective in a package of three is 0.0294 or approximately 2.94%.