Question

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are rated 23-watt. Suppose that three of these bulbs are randomly selected. (Round your answers to three decimal places.) (a) What is the probability that exactly two of the selected bulbs are rated 23-watt? (b) What is the probability that all three of the bulbs have the same rating? (c) What is the probability that one bulb of each type is selected?

Answer 1

(a) The probability that exactly two of the selected bulbs are rated 23-watt is 0.259. (b) The probability that all three of the bulbs have the same rating is 0.021. (c) The probability that one bulb of each type is selected is 0.333.

Explanation:

In order to find the probability of an event, we first need to determine the total number of possible outcomes and the number of favorable outcomes. In this case, there are 24 bulbs in the box and we are selecting 3 of them, which means there are 24C3 = 2024 possible outcomes.

(a) To find the probability that exactly two of the selected bulbs are rated 23-watt, we need to determine the number of favorable outcomes. Since there are 7 bulbs rated 23-watt, the first selected bulb can be any of those 7, the second selected bulb can also be any of the 7, and the third selected bulb must be one of the remaining 17 bulbs. Therefore, the number of favorable outcomes is 7C2 * 17C1 = 7 * 17 = 119. Hence, the probability is 119/2024 = 0.259.

(b) To find the probability that all three of the bulbs have the same rating, we need to determine the number of favorable outcomes. Since there are 3 different ratings, the first selected bulb can be any of the 24 bulbs, the second selected bulb must be one of the remaining bulbs with the same rating, and the third selected bulb must be one of the remaining bulbs with the same rating. Therefore, the number of favorable outcomes is 24 * 8 * 3 = 576. Hence, the probability is 576/2024 = 0.021.

(c) To find the probability that one bulb of each type is selected, we need to determine the number of favorable outcomes. Since there are 8, 9, and 7 bulbs of each type, the first selected bulb can be any of those numbers, the second selected bulb must be one of the remaining bulbs of a different type, and the third selected bulb must be one of the remaining bulbs of the last type. Therefore, the number of favorable outcomes is 8 * 9 * 7 = 504. Hence, the probability is 504/2024 = 0.333.

In conclusion, the probabilities for (a), (b), and (c) are 0.259, 0.021, and 0.333 respectively. These probabilities can also be verified by adding up the individual probabilities of each event, which should equal 1. This type of problem is commonly used in statistics and probability to determine the likelihood of an event occurring. By understanding and applying the concepts of combinations and probabilities, we can accurately calculate the chances of various outcomes and make informed decisions.

Answer 2

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

`P = p^(3)_(2,1)*(7)/(24)*(6)/(23)*(17)/(22) = (3!)/(2!1!)*(7)/(24)*(6)/(23)*(17)/(22) = 3*(7)/(24)*(6)/(23)*(17)/(22) = 0.1764`

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

`P = P_(1) + P_(2) + P_(3)`

`P_(1)` is the probability that all three of them are 13-watt. So:

`P_(1) = (8)/(24)*(7)/(23)*(6)/(22) = 0.0277`

`P_(2)` is the probability that all three of them are 18-watt. So:

`P_(2) = (9)/(24)*(8)/(23)*(7)/(22) = 0.0415`

`P_(3)` is the probability that all three of them are 23-watt. So:

`P_(3) = (7)/(24)*(6)/(23)*(5)/(22) = 0.0173`

`P = P_(1) + P_(2) + P_(3) = 0.0277 + 0.0415 + 0.0173 = 0.0865`

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

`P = p^(3)_(1,1,1)*(8)/(24)*(9)/(23)*(7)/(22) = 3**(8)/(24)*(9)/(23)*(7)/(22) = 0.1245`

There is a 12.45% probability that one bulb of each type is selected.