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

Answer:

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.

Step-by-step 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

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$

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:

$P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277$

$P_{2}$ is the probability that all three of them are 18-watt. So:

$P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415$

$P_{3}$ is the probability that all three of them are 23-watt. So:

$P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173$

$P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865$

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

$P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245$

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