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.)

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!

after simplifying it i got: -36.25 as my answer. hope that !

answer: 2x + 2 − 2 = 10 − 2; subtraction property of equality

step-by-step explanation: