Question

A box in a certain supply room contains 540−W lightbulbs, 460−W lightbulbs, and 675−W bulbs. Suppose that 3 bulbs are randomly selected without replacement. (Round your answers to 4 decimal places, if needed.) (a) What is the probability that exactly two of the selected bulbs are rated 75−W ? (b) What is the probability that all three of the selected bulbs have the same rating? (c) What is the probability that one bulb of each type is selected? (d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?

Answer 1

Let's solve each part of this problem step by step:

(a) Probability that exactly two of the selected bulbs are rated 75-W:

First, we need to calculate the probability of selecting two 75-W bulbs and one bulb of a different rating. There are three ways to choose which bulb is not rated 75-W (either 540-W, 460-W, or 675-W), so we'll calculate the probability for each case and then sum them up.

Case 1: Two 75-W bulbs and one 540-W bulb:

Probability = (C(3, 2) * C(540, 2) * C(460, 1)) / C(1575, 3)

Case 2: Two 75-W bulbs and one 460-W bulb:

Probability = (C(3, 2) * C(540, 2) * C(675, 1)) / C(1575, 3)

Case 3: Two 75-W bulbs and one 675-W bulb:

Probability = (C(3, 2) * C(460, 2) * C(540, 1)) / C(1575, 3)

Now, calculate the probabilities for each case:

Case 1: (3 choose 2) * (540 choose 2) * (460 choose 1) / (1575 choose 3) ≈ 0.0592

Case 2: (3 choose 2) * (540 choose 2) * (675 choose 1) / (1575 choose 3) ≈ 0.0544

Case 3: (3 choose 2) * (460 choose 2) * (540 choose 1) / (1575 choose 3) ≈ 0.0343

Now, sum up these probabilities to get the overall probability:

Total probability = Case 1 + Case 2 + Case 3 ≈ 0.0592 + 0.0544 + 0.0343 ≈ 0.1480

(b) Probability that all three of the selected bulbs have the same rating:

There are three possibilities for this scenario: all three bulbs are 540-W, all three bulbs are 460-W, or all three bulbs are 675-W.

Probability = (C(540, 3) + C(460, 3) + C(675, 3)) / C(1575, 3)

Calculate each case and then sum them up:

Probability = (C(540, 3) + C(460, 3) + C(675, 3)) / (1575 choose 3) ≈ (0.0071 + 0.0051 + 0.0159) ≈ 0.0281

(c) Probability that one bulb of each type is selected:

To calculate this probability, we need to consider all possible combinations of selecting one bulb of each type (75-W, 540-W, and 460-W).

Probability = (C(3, 1) * C(3, 1) * C(3, 1)) / C(1575, 3)

Calculate:

Probability = (3 choose 1) * (3 choose 1) * (3 choose 1) / (1575 choose 3) ≈ 0.0045

(d) Probability that it is necessary to examine at least six bulbs to find a 75-W bulb:

The probability of finding a 75-W bulb in the first five attempts is the complement of the probability of finding a 75-W bulb within the first five attempts.

P(Finding a 75-W bulb within 5 attempts) = 1 - P(Not finding a 75-W bulb in 5 attempts)

The probability of not finding a 75-W bulb in one attempt is the sum of the probabilities of selecting a 540-W bulb, a 460-W bulb, and a 675-W bulb.

P(Not finding a 75-W bulb in one attempt) = (C(540, 1) + C(460, 1) + C(675, 1)) / (C(1575, 1))

Now, calculate the probability of not finding a 75-W bulb in five attempts:

P(Not finding a 75-W bulb in 5 attempts) = (P(Not finding a 75-W bulb in one attempt))^5

Then, calculate the final probability:

P(Finding a 75-W bulb within 5 attempts) = 1 - P(Not finding a 75-W bulb in 5 attempts)

Calculate each step, and you will get the answer.

Explanation: