Question

A box in a supply room contains 22 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 5 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?

Answer 1

a)P(exactly 2 bulb from watt)= 0.11

b)P=0.097

c)P=0.233

d)P=0.235

Step-by-step explanation:

Total bulb = 22

13 watt bulb = 8

18 watt = 9

23 watt = 5

We have to select 3 bulb out of 22 bulb

We know that probability is the ratio of favor condition to the total condtion.

The total number of way to select 3 bulb out of 22 =N

`N=_(3)^(22)\textrm{c}`

a)

`P(exactly\ 2\ bulb\ from\ 23\ watt)= \frac{_(2)^(5)\textrm{c}._(1)^(17)\textrm{c}}{_(3)^(22)\textrm{c}}`

`P(exactly\ 2\ bulb\ from\ 23\ watt)= (170)/(1540)`

P(exactly 2 bulb from watt)= 0.11

b)

The probability that all three of the bulbs have the same rating =P

`P=\frac{_(3)^(5)\textrm{c}+_(3)^(8)\textrm{c}+_(3)^(9)\textrm{c}}{_(3)^(22)\textrm{c}}`

`P= (150)/(1540)`

P=0.097

c)

The probability that one bulb of each type is selected=P

`P=\frac{_(1)^(5)\textrm{c}._(1)^(8)\textrm{c}._(1)^(9)\textrm{c}}{_(3)^(22)\textrm{c}}`

`P= (360)/(1540)`

P=0.233

d)

The probability that it is necessary to examine at least 6 bulbs =P

`P=\frac{_(5)^(17)\textrm{c}}{_(5)^(22)\textrm{c}}`

`P= (6188)/(26334)`

P=0.235