(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
