Answer:
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