Question

Each of 12 refrigerators of a certain type has a 10% chance of needing repairs in a given year. What is the probability that more than one refrigerator will need repairs in a year?Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let X be the number among the first 6 examined that have a defective compressor.

a.) Calculate P(X=4) and P(X≤4)

Answer 1

The probability that more than one refrigerator will need repairs in a year is approximately 0.421. This can be calculated using the binomial probability formula.

Step-by-step explanation:

To calculate the probability that more than one refrigerator will need repairs in a year, we need to calculate the probability that exactly one refrigerator will need repairs and subtract it from 1. We can use the binomial probability formula to calculate this.

The probability of each refrigerator needing repairs is 0.10. Let X be the number of refrigerators that need repairs in a year. Since there are 12 refrigerators, X follows a binomial distribution with parameters n = 12 and p = 0.10. We need to calculate P(X > 1) = 1 - P(X ≤ 1).

P(X = 1) = C(12, 1) * (0.10)^1 * (1 - 0.10)^(12 - 1) = 12 * 0.10 * 0.90^11 ≈ 0.297

P(X > 1) = 1 - P(X ≤ 1) = 1 - [P(X = 0) + P(X = 1)] ≈ 1 - [C(12, 0) * (0.10)^0 * (1 - 0.10)^(12 - 0) + P(X = 1)] ≈ 1 - [1 * 1 * 0.90^12 + 0.297] ≈ 1 - [0.282 + 0.297] ≈ 1 - 0.579 ≈ 0.421