- #1

jaus tail

- 615

- 48

## Homework Statement

In a company, out of a sample of 20 bulbs, the mean number of defective bulbs are 2.

Out of 1000 such samples, how many samples would have atleast 3 defective bulbs?

## Homework Equations

Mean = n * p where n is number of bulbs and p is probability of a bulb being defective.

P(x succeed at r times) when the experiment is done n times = nCr p

^{r}q

^{(n-r)}

where p is probability success when experiment is done once

and q = 1 - p

## The Attempt at a Solution

Mean number of defectives = 2 = n * p. n = 20.

so p =2/20 = 0.1. Probability of bulb being defective

and q = 1 - 0.1 = 0.9. Probability of bulb not being defective.

Probability of at least 3 defective in a sample of 20

= 1 - [ P(zero defective) + P(1 defective) + (P(2 defective) ]

Now n = 20. r = 0 here r =1 here r = 2 here

p = 0.1

q = 0.9

P(atleast 3 in sample of 20) = 1 - [

^{20}C

_{0}(0.1)

^{0}(0.9)

^{20}+

^{20}C

_{1}(0.1)

^{1}(0.9)

^{19}+

^{20}C

_{2}(0.1)

^{2}(0.9)

^{18})

*This i get as 0.323.*

So the probability of at least three defective in a sample of 20 bulbs is 0.323.

So the probability of at least three defective in a sample of 20 bulbs is 0.323.

Now how to proceed?

I have to find 'How many such samples would be expected to contain atleast 3 defective bulbs'?

Answer is 323. I guess they've multplied 0.323 by number of samples but why?PS... sorry for too many homework posts. Have an exam coming on 11th Feb.