A box in a supply room contains 20 compact fluorescent light

A box in a supply room contains 20 compact fluorescent lightbulbs, of which 7 are rated 13-watt, 9 are rated 18-watt, and 4 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?


(d) If bulbs are selected one by one until a 23-watt bulb is obtained, what is the probability that it is necessary to examine at least 6 bulbs?

Solution

A box in a supply room contains 20 compact fluorescent lightbulbs, of which 7 are rated 13-watt, 9 are rated 18-watt, and 4 are rated 23-watt.

total number of balls = 7 + 9 + 4 = 20

We have to select 3 balls from 20.

Here we use combination formula,

nCr = n! / r!*(n-r)!

This we can done by using 20 C 3 ways where C is used for combination.

20C3 = 20! / 3!*17! = 1140

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

P(exactly two of the selected bulbs are rated 23-watt) =

Different ways to pull two 23 watt bulbs out of the 4 that are there is (4 C 2).

4C2 = 4! / 2!*2! = 6

Different ways to pull one non-23 watt bulb out of the 20-4 =16 that are there is (16 C 1).

16C1 = 16! / 15!*1! = 16

P(exactly two of the selected bulbs are rated 23-watt) = (6*16) / 1140 = 0.084

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

Different ways to pull three 13 watt bulbs out of the 7 that are there is 7C3.

7C3 = 7! / 3!/4! = 35

Different ways to pull three 18 watt bulbs out of the 9 that are there is 9C3.

9C3 = 9! / 3!*6! = 84

Different ways to pull three 23 watt bulbs out of the 4 that are there is 4C3.

4C3 = 4! / 3!*1! = 4

P( all three of the bulbs have the same rating) = (35+84+4) / 1140 = 123/1140 = 0.108

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

Different ways to pull one 13 watt bulb out of the 7 that are there is 7C1.

7C1 = 7! / 1!*6! = 7

Different ways to pull one 18 watt bulb out of the 9 that are there is 9C1.

9C1 = 9! / 1!*8! = 9

Different ways to pull one 23 watt bulb out of the 4 that are there is 4C1.

4C1 = 4! / 1!*3! = 4

P(one bulb of each type is selected) = (7*9*4) / 1140 = 0.221

(d) If bulbs are selected one by one until a 23-watt bulb is obtained, what is the probability that it is necessary to examine at least 6 bulbs?

Chances of finding it in exactly one try = 4/20 = 0.2

Chances of finding it in exactly two tries = 16/20 * 4/19 = 0.168

Chances of finding it in exactly three tries = 16/20 * 15/19 * 4/18 = 0.140

Chances of finding it in exactly four tries = 16/20 * 15/19 * 14/18 * 4/17 = 0.116

Chances of finding it in exactly five tries = 16/20 * 15/19 * 14/18 * 13/17 * 4/16 = 0.094

Add all these together... = 0.2 + 0.168 + 0.140 + 0.116 + 0.094 = 0.718

This is the probability that you WILL select a 23-watt bulb in the first five tries.

So, the chances that it will take \"at least 6\" tries to find it is 1 - 0.718 = 0.282