# Thread: Stuck on simple probability question

1. ## Stuck on simple probability question

Hello all,

I've been stuck for the last couple of days on this supposedly very simple probably questions:

Question:
30 widgets have been manufactured. 5 of which are known to be faulty. If 5 were randomly selected, what is the probability that exactly 3 of the 5 chosen will be faulty?

There are a total of 30-choose-5 unique combinations, I've figured out that only 3000 of these combinations contains only 3 widgets that are faulty (aka ignored combinations where 4 and 5 widgets were faulty). This gives a probability of 3000/142506 0r roughly 0.02105

I was wondering how does one figure this out using probability theory, as the above value of 3000 was found using a computer program that went through all the 30-choose-5 combinations.

Furthermore I don't think this is a binomial probability problem, as the total number of widgets is known, and the act of replacing does not occur. Assistance will be greatly appreciated.

2. Originally Posted by Sherrie
Question:
30 widgets have been manufactured. 5 of which are known to be faulty. If 5 were randomly selected, what is the probability that exactly 3 of the 5 chosen will be faulty?
I was wondering how does one figure this out using probability theory, as the above value of 3000 was found using a computer program that went through all the 30-choose-5 combinations.
This is a simple combination application.
$\dfrac{\binom{5}{3}\binom{25}{2}}{\binom{30}{5}}$.

Where $\dbinom{N}{K}=\dfrac{N!}{K!(N-K)!}$

3. Hello Plato,

Thank-you for the quick response!

I was wondering if you could explain a bit more about where 25-choose-2 came from and how its being used. I figure the 25 is the result of the remainder of chooseing 5 lots from 30, but in plain terms I don't quite understand how it all comes together.

For example does the logic change if say 6 widgets were faulty instead of 5, yet 5 are still randomly chosen? would it be (6-choose-3) * (24-choose-2) / (30-choose-5) ?

btw is there a specific name or term used for the above solution? (its not binomial probability i think.)

4. Originally Posted by Sherrie
I was wondering if you could explain a bit more about where 25-choose-2 came from and how its being used. I figure the 25 is the result of the remainder of chooseing 5 lots from 30, but in plain terms I don't quite understand how it all comes together.
btw is there a specific name or term used for the above solution?
To get exactly three defective items in a total of five, we pick three from the defective items and two from the twenty-five non-defective items.

This called sampling without replacement.

5. Originally Posted by Plato
To get exactly three defective items in a total of five, we pick three from the defective items and two from the twenty-five non-defective items.
I see, so for every one of the 5-choose-3 combinations there are 25-choose-2 combinations, which explains your original solutions. thanks!