Question

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

Answer 1

Answer:a) Probability P(exactly 2 bulbs rated 13watts)= 0.22

b) Probability(each bulb different rating)

= 0.24

Step-by-step explanation:

There are 6 13watts bulbs

8 18watts bulbs

4 23watts bulbs

Total bulbs = 18

a) Probability that all 3 bulbs are 18watts

Number of ways of pulling 3 bulbs = 18!/(3!×15!) = (6.4×10^15)/7.846×10^12) = 815 ways

Different ways of pulling 13watts bulbs out of 6 = 6!/(2!×4!)= 720/48=15ways

Different ways of pulling non 13 watts bulbs= 12!/(1!×11!) = 479,001,600/ 39,916,800 = 12

Number of ways total= 15×12=180waya

Therefore P(exactly 2 bulbs rated 13watts)= 180/815 =0.22

b) Probability P( all 3 bulbs are 1 from each rating)

Ways of pulling 3bulbs bulb each from 3 ratings are 4× 5 × 8= 192ways

Probability = 192/815 =0.24

Answer 2

Answer:

a) 0.103

b) 0.098

c) 0.235

d) 0.054

Step-by-step explanation:

Total number of bulbs = 18

Total number of 13-watt bulbs = 6

Total number of 18-watt bulbs = 8

Total number of 23-watt bulbs = 4

Let A represent 13-watt bulbs, B represent 18-watt bulbs and C represent 23-watt bulbs.

a) Probability that exactly two of the bulbs are rated 23-watts with order not important = P(C n C n (AuB)) + P(C n (AuB) n C) + ((AuB) n C n C) = ((4/18) × (3/17) × (14/16)) + ((4/18) × (14/17) × (3/16)) + ((14/18) × (4/17) × (3/16)) = 0.103 to 3d.p

b) Probability that all 3 bulbs have the same rating = P(A n A n A) + P(B n B n B) + P(B n B n B) = (6/18 × 5/17 × 4/16) + (8/18 × 7/17 × 6/16) + (4/18 × 3/17 × 2/16) = 0.0245 + 0.0686 + 0.0049 = 0.098 to 3d.p

c) probability that one bulb of each type is selected = P(A n B n C) + P(A n C n B) + P(B n A n C) + P(B n C n A) + P(C n A n B) + P(C n B n A) = 6(P(A n B n C)) = 6 (6/18 × 8/17 × 4/16) = 6 × 0.0392 = 0.235 to 3dp

d) probability that it is necessary to examine at least 6 bulbs if bulbs are selected one by one until a 23-watt bulb is obtained

To do this, we'll group the 13 and 18 watt bulbs as one, (AuB) = A*. n(A*) = 14

Our probability = P(A* × A* × A* × A* × A* × A* × C) = (14/18 × 13/17 × 12/16 × 11/15 × 10/14 × 9/13 × 4/12) = 0.0539 = 0.054 to 3dp

Hope this helps!