A box in a supply room contains 23 compact fluorescent light

A box in a supply room contains 23 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 6 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 23 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 6 are rated 23-watt.

total number of balls = 8 + 9 + 6 = 23

We have to select 3 balls from 23.

Here we use combination formula,

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

This we can done by using 23 C 3 ways where C is used for combination.

23C3 = 23! / 3!*20! = 1771

(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 6 that are there is (6 C 2).

6C2 = 6! / 2!*4! = 15

Different ways to pull one non-23 watt bulb out of the 23-6 =17 that are there is (17 C 1).

17C1 = 17! / 16!*1! = 17

P(exactly two of the selected bulbs are rated 23-watt) = (15*17) / 1771 = 0.144

(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 8 that are there is 8C3.

8C3 = 8! / 3!/5! = 56

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 6 that are there is 6C3.

6C3 = 6! / 3!*3! = 20

P( all three of the bulbs have the same rating) = (56+84+20) / 1771 = 160/1771 = 0.090

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

Different ways to pull one 13 watt bulb out of the 8 that are there is 8C1.

8C1 = 8! / 1!*7! = 8

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 6 that are there is 6C1.

6C1 = 6! / 1!*5! = 6

P(one bulb of each type is selected) = (8*9*6) / 1771 = 0.244

(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 = 6/23 = 0.261

Chances of finding it in exactly two tries = 17/23 * 6/22 = 0.202

Chances of finding it in exactly three tries = 17/23 * 16/22 * 4/21 = 0.102

Chances of finding it in exactly four tries = 17/23 * 16/22 * 15/21 * 4/ 20 = 0.077

Chances of finding it in exactly five tries = 17/23 * 16/22 * 15/21 * 14/20 * 4/19 = 0.057

Add all these together... = 0.261 + 0.202 + 0.102 + 0.077 + 0.057 = 0.698

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 - 0.698= 0.302