Consider three boxes containing a brand of light bulbs. Box I contains 6 bulbs of which 2 are defective, Box 2 has 1 defective and 3 functional bulbs and Box 3
contains 3 defective and 4 functional bulbs. A box is selected at random and a bulb
drawn from it at random is found to be defective. Find the probability that the box
selected was Box 2.

1/6

Step-by-step explanation:

As we already know that selected bulb is defective the required probability doesn't depend on functional bulbs at all.

The probability, that selected defective bulb is from Box2 is number of defective bulbs in Box 2  divided by total number of defective bulbs.

P(defective in box 2)= N(defective in box 2)/N(defective total)

As we know there is only 1 defective lamp in box 2.

So N(defective in box 2)=1

Total number of defective bulbs is Box1- 2 defective  bulbs,  box2- 1 defective bulbs, box3 - 3 defective bulbs. Total are 6 defective bulbs.

So N(defective total)=6

So P(defective in box 2)=1/6

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6

the answer is b.

step-by-step explanation: