Question

a manufacturer of sells light bulbs in boxes of 8. any given light bulb has a 5% probability of being defective. what is the probability that in a box of eight, two or more bulbs will be defective?

Answer 1

Explanation:

Step 1: Identify the given information

- Number of light bulbs in a box (n) = 8

- Probability of a light bulb being defective (p) = 5% = 0.05

- Probability of a light bulb not being defective (q) = 1 - p = 1 - 0.05 = 0.95

Step 2: Calculate the probability of having exactly 0, 1, or 2 defective light bulbs in a box

We will use the binomial probability formula: P(x) = C(n, x) * p^x * q^(n-x)

For 0 defective light bulbs:

P(0) = C(8, 0) * (0.05)^0 * (0.95)^8

P(0) = 1 * 1 * 0.6634

P(0) = 0.6634

For 1 defective light bulb:

P(1) = C(8, 1) * (0.05)^1 * (0.95)^7

P(1) = 8 * 0.05 * 0.6983

P(1) = 0.2793

Step 3: Calculate the probability of having 2 or more defective light bulbs

Since the probabilities of all possible outcomes must add up to 1, we can find the probability of having 2 or more defective light bulbs by subtracting the probabilities of having 0 or 1 defective light bulbs from 1.

P(2 or more) = 1 - P(0) - P(1)

P(2 or more) = 1 - 0.6634 - 0.2793

P(2 or more) = 0.0573