Defective light bulbs - Joint probability

[hadroneater]

Homework Statement

Of 25 light bulbs, 5 are defective. 8 bulbs are chosen from the 25. What is the probability that 2 or fewer of the 8 selected light bulbs are defective?

Homework Equations

The Attempt at a Solution

It's been awhile since I've done questions like this one. I was thinking it has something to do with permutations or perhaps some sort of random variable distribution?

 

[Dick]

Well, one way to do it is to chose 2 of the defective bulbs and 6 of the good bulbs. Or you could choose 1 defective and 7 good. Or just choose 8 good. Do you know how to count the ways for those cases? Then just divide the total number of ways to choose 8 bulbs.

 

[quincyboy7]

Homework Statement

Of 25 light bulbs, 5 are defective. 8 bulbs are chosen from the 25. What is the probability that 2 or fewer of the 8 selected light bulbs are defective?

Let's go through some casework.

If no selected light bulbs are defective, there is a (4/5)^8 chance of that happening

If 1 selected light bulb is defective, there is a (4/5)^7*(1/5)*8 chance of that happening.

If 2 selected light bulbs are defective, there is a (4/5)^6*(1/5)^2*8*7 chance of that happening. How I got the integers is by doing 8C8=1, 8C7=8, and 8C6=8*7. This is because order does not matter i.e. if you pick a defective then a good one, it's the same as picking a good one, then a defective.

 

[Ray Vickson]

Let's go through some casework.

If no selected light bulbs are defective, there is a (4/5)^8 chance of that happening

If 1 selected light bulb is defective, there is a (4/5)^7*(1/5)*8 chance of that happening.

If 2 selected light bulbs are defective, there is a (4/5)^6*(1/5)^2*8*7 chance of that happening. How I got the integers is by doing 8C8=1, 8C7=8, and 8C6=8*7. This is because order does not matter i.e. if you pick a defective then a good one, it's the same as picking a good one, then a defective.

Your argument applies to sampling "with replacement", where each successive drawing is independent of previous drawings (essentially because we put back the bulb after drawing it out). That is not normally how sampling is done in practice. Instead, we should probably look at sampling "without replacement". This would give P{no defects in sample} = (20/25)*(19/24)*(18/23)*...*(13/18), because after removing the first non-defective we are left with 24 bulbs, of which 19 are non-defective, etc.

RGV