Question

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.
Find the probability that this box will contain at most 20 defective light bulbs. Show your work or calculator input.

Answer 1

Answer:

Total probability = 0.9278

Step-by-step explanation:

Probability one of the bulbs is defective = 0.1

Probability of one bulb being defective = $0.1*0.9^1^4^9$  * Number of possible arrangements

Probability of one bulb being defective = $0.1*0.9^1^4^9 * 150$

Probability of one bulb being defective = 0.00000228

Similarly, the probability of 2 to 20 bulbs being defective can be found. We simply need to find the probability of the event happening and multiply it with the number of ways we can arrange the bulbs.

Let X be the number of defective bulbs. we then get the following equations:

Probability of event happening = 0.1 ^ (X) * 0.9 (150-X)

Arrangement (number of ways it could happen) = 150! / [ X! * (150-X)! ]

The total probability will then be the above equations multiplied. Given below are the answers for each type of event:

2 defective bulbs = 0.00001885

3 defective bulbs = 0.0001035

4 defective bulbs = 0.0004227

5 defective bulbs = 0.001371

6 defective bulbs = 0.003682

7 defective bulbs = 0.008417

8 defective bulbs = 0.01672

9 defective bulbs = 0.02931

10 defective bulbs = 0.04591

11 defective bulbs = 0.06493

12 defective bulbs = 0.08357

13 defective bulbs = 0.09857

14 defective bulbs = 0.10717

15 defective bulbs = 0.1079

16 defective bulbs = 0.1012

17 defective bulbs = 0.08865

18 defective bulbs = 0.07278

19 defective bulbs = 0.05618

20 defective bulbs = 0.04088

Adding up all the above probabilities we get the answer = 0.9278