Probability Bionomial distribution

[jaus tail]

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^rq^(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 - [ ²⁰C₀ (0.1)⁰(0.9)²⁰ + ²⁰C₁(0.1)¹ (0.9)¹⁹ + ²⁰C₂(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.

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.

 

[Ray Vickson]

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^rq^(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 - [ ²⁰C₀ (0.1)⁰(0.9)²⁰ + ²⁰C₁(0.1)¹ (0.9)¹⁹ + ²⁰C₂(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.

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.

They computed the expected (or mean) number of samples having at least 3 defective. That is what their very poorly-worded problem actually wanted.

 

[Stephen Tashi]

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

That part of the problem is badly written. We can not determine how many out of 1000 samples would definitely have at least 3 defective bulbs. An answerable question would be "Out of 100 such samples, how many of them, on the average would contain at least 3 defective bulbs". Assuming that is the correct interpretation, you can justify the answer by using the formula for the mean of a binomial distribution.( There is also some ambiguity in the phrase "such samples". That could be interpreted to refer to any sample of 20 bulbs or it could be interpreted to mean any sample of 20 bulbs that had 2 defective bulbs. Presumably, the intended meaning is "any sample of 20 bulbs".)

 

[jaus tail]

Oh so they've used formula mean = n * p

first n = 20. and mean is 2. so Probability of 1 defective bulb = 0.1
p = (probability of 1 bulb defective out of 20)

using this we find probability of at least 3 defective bulbs out of 20, using Bionomial distribution.

Then new P is of atleast three defective bulbs out of 20.
Now n changes from 20 to 1000.

Wow. this sure is confusing.

 

[jaus tail]

Well the question 'word to word' is:
In sampling a large number of parts manufactured by a machine, the mean number of defectives in a sample of 20 is 2. Out of 1000 such samples, how many would be expected to contain at least 3 defective parts.

 

[Ray Vickson]

Well the question 'word to word' is:
In sampling a large number of parts manufactured by a machine, the mean number of defectives in a sample of 20 is 2. Out of 1000 such samples, how many would be expected to contain at least 3 defective parts.

You should have written this in the beginning; it is so much clearer and so much more accurate than the version you posted. For one thing, saying "In sampling... the mean number is ..." makes perfect probabilistic sense; the version you wrote does not. Secondly, the word "expected" in the second sentence is crucial, and should never have been omitted.

 

[jaus tail]

Oh... sorry for that. It's confusing when learning probability and statistics together. Had to go through links to understand why expected value and mean should be same.

Thanks again.

 

[Ray Vickson]

Oh... sorry for that. It's confusing when learning probability and statistics together. Had to go through links to understand why expected value and mean should be same.

Thanks again.

Actually, in Statistics, the mean and the expectation are not always the same thing! Typically, mean refers to the average of some data, while expected value refers to the first-moment of some underlying probability distribution. For example, the mean of the numbers {1,2,3} is (1+2+3)/3 = 2, but all three could be sample-points from a probability distribution having expected value 1.5.

In probability alone (i.e., outside of Statistics) the words mean and expectation are often used interchangeably, although some would say that is sloppy language.

Anyway, in this problem you have two applications of the binomial distribution: one (the [20, 0.10] one) is used to compute the appropriate probability parameter $p$ for the second binomial (the [1000,p] one).