Abox 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 ) 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?

(a) 0.110 ⇒ rounded to three decimal places

(b) 0.097 ⇒ rounded to three decimal places

(c) 0.234 ⇒ rounded to three decimal places

(d) 0.235 ⇒ rounded to three decimal places

Step-by-step explanation:

* Lets solve the problem using the combination

- The order is not important in this problem, so we can use the

  combination nCr to find the probability

- There are 22 compact fluorescent light bulbs

# 8 ⇒ rated 13 watt

# 9 ⇒ rated 18 watt

# 5 ⇒ rated 23 watt

- 3 bulbs are randomly selected

(a) Exactly two of the selected bulbs are rated 23 watt

∵ Exactly two of them are rated 23 watt

∵ The number of bulbs rated 23 watt is 5

∴ We will chose 2 from 5

∴ 5C2 = 10 ⇒ by using calculator or by next rule

# nCr = n!/[r! × (n-r)!]

- 5C2= 5!/[2! × (5 - 2)!] = (5×4×3×2×1)/[(2×1)×(3×2×1)] = 120/12 = 10

∴ There are 10 ways to chose 2 bulbs from 5

∵ The 3rd bulbs will chosen from the other two types

∵ The other two types = 8 + 9 = 17

∴ We will chose 1 bulbs from 17 means 17C1

∵ 17C1 = 17 ⇒ (by the same way above)

∴ There are 17 ways to chose 1 bulbs from 17

- 10 ways for two bulbs and 17 ways for one bulb

∴ There are 10 × 17 = 170 ways two chose 3 bulbs exact 2 of them

   rated 23 watt (and means multiply)

∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs

∵ 22C3 = 1540 ⇒ (by the same way above)

∴ The probability = 170/1540 = 17/154 = 0.110

* The probability that exactly two of the selected bulbs are rated

  23-watt is 0.110 ⇒ rounded to three decimal places

(b) All three of the bulbs have the same rating

- We can have either all 13 watt, all 18 watt or all 23 watt.

# 13 watt

∵ There are 8 are rated 13 watt

∵ We will chose 3 of them

∴ There are 8C3 ways to chose 3 bulbs from 8 bulbs

∵ 8C3 = 56

∴ There are 56 ways to chose 3 bulbs rated 13 watt

# 18 watt

∵ There are 9 are rated 18 watt

∵ We will chose 3 of them

∴ There are 9C3 ways to chose 3 bulbs from 9 bulbs

∵ 9C3 = 84

∴ There are 84 ways to chose 3 bulbs rated 18 watt

# 23 watt

∵ There are 5 are rated 23 watt

∵ We will chose 3 of them

∴ There are 5C3 ways to chose 3 bulbs from 5 bulbs

∵ 5C3 = 10

∴ There are 10 ways to chose 3 bulbs rated 23 watt

- 65 ways or 84 ways or 10 ways for all three of the bulbs have

 the same rating

∴ There are 56 + 84 + 10 = 150 ways two chose 3 bulbs have the

   same rating (or means add)

∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs

∵ 22C3 = 1540

∴ The probability = 150/1540 = 15/154 = 0.097

* The probability that all three of the bulbs have the same rating is

  0.097 ⇒ rounded to three decimal places

(c) One bulb of each type is selected

- We have to select one bulb of each type

# 13 watt

∵ There are 8 bulbs rated 13 watt

∵ We will chose 1 of them

∴ There are 8C1 ways to chose 1 bulbs from 8 bulbs

∵ 8C1 = 8

∴ There are 8 ways to chose 1 bulbs rated 13 watt

# 18 watt

∵ There are 9 bulbs rated 18 watt

∵ We will chose 1 of them

∴ There are 9C1 ways to chose 1 bulbs from 9 bulbs

∵ 9C1 = 9

∴ There are 9 ways to chose 1 bulbs rated 18 watt

# 23 watt

∵ There are 5 bulbs rated 23 watt

∵ We will chose 1 of them

∴ There are 5C1 ways to chose 1 bulbs from 5 bulbs

∵ 5C1 = 5

∴ There are 5 ways to chose 1 bulbs rated 23 watt

- 8 ways and 9 ways and 5 ways for one bulb of each type is selected

∴ There are 8 × 9 ×  5 = 360 ways that one bulb of each type is

  selected (and means multiply)

∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs

∵ 22C3 = 1540

∴ The probability = 360/1540 = 18/77 = 0.234

* The probability that one bulb of each type is selected is 0.234

  ⇒ rounded to three decimal places

(d) If bulbs are selected one by one until a 23-watt bulb is obtained,

    it is necessary to examine at least 6 bulbs

- We have a total of 22 bulbs and 5 of them are rate 23 watt

∴ There are 22 - 5 = 17 bulbs not rate 23 watt

∵ We examine at least 6 so the 6th one will be 23 watt

∴ There are 17C5 ways that the bulb not rate 23 watt

∵ 17C5 = 6188

∴ There are 6188 ways that the bulb is not rate 23 watt

∵ There are 22C5 ways to chose 5 bulbs from total 22 bulbs

∵ 22C5 = 26334

∴ The probability = 6188/26334 = 442/1881 = 0.235

* If bulbs are selected one by one until a 23-watt bulb is obtained,

 the probability that it is necessary to examine at least 6 bulbs is

  0.235 ⇒ rounded to three decimal places

OR

we can use the rule:

# P ( examine at least six ) = 1 − P ( examine at most five )  

# P ( examine at least six ) = 1 − P (1) − P (2) − P (3) − P (4) − P (5)

  where P is the probability  

∵ P(1) = 5/22

∵ P(2) = (5 × 17)/(22 × 21) = 85/462

∵ P(3) = (5 × 17× 16)/(22 ×21 × 20) = 34/231

∵ P(4) = (5 × 17 × 16 × 15)/(22 ×21 × 20 ×19) = 170/1463

∵ P(5) = (5 × 17 × 16 × 15 × 14)/(22 × 21 × 20 × 19 × 18) = 170/1881

∴ P = 1 - 5/22 - 85/462 - 34/231 - 170/1463 - 170/1881 = 442/1881

∴ P = 0.235  

-6 is the correct option from the above given options