# Defective light bulbs - Joint probability

• hadroneater
In summary: There are a total of 18 ways to select 8 members from that set. So the probability of having exactly 2 or fewer defective bulbs in the sample is 18/18 or 1.
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?

## 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?

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.

hadroneater said:

## 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.

quincyboy7 said:
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

## 1. What is the definition of a defective light bulb?

A defective light bulb is a light bulb that does not function properly, either due to damage or a manufacturing error.

## 2. What is the joint probability of a light bulb being defective?

The joint probability of a light bulb being defective is the probability that both the light bulb is defective and it is selected from a given sample size of light bulbs.

## 3. How is the joint probability calculated for defective light bulbs?

The joint probability of defective light bulbs is calculated by dividing the number of defective light bulbs by the total number of light bulbs in the sample.

## 4. How does the joint probability of defective light bulbs affect the overall reliability of light bulbs?

The joint probability of defective light bulbs is an important factor in determining the overall reliability of light bulbs. A higher joint probability indicates a higher likelihood of a light bulb being defective, which can affect the reliability and performance of the light bulb.

## 5. Can the joint probability of defective light bulbs be reduced?

The joint probability of defective light bulbs can be reduced through careful quality control measures in the manufacturing process and thorough testing of light bulbs before they are distributed to consumers.

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