Probability of Defective Fluorescent Light Bulbs in a Sample - Analyzing and Solving with Binomial Distribution

Introduction

When dealing with a factory that manufactures fluorescent light bulbs and finds that 10 out of every 100 bulbs are defective, it becomes essential to understand the probability of finding defective bulbs in a random sample. This problem is often approached using the binomial distribution, which models the number of successes in a fixed number of independent trials. Here, we will explore the steps to calculate the probability of having a certain number of defective bulbs in a sample of 6 using binomial distribution.

Understanding the Problem

The problem states that 10 out of 100 fluorescent light bulbs manufactured by a factory are defective. We are asked to calculate the probability that out of 6 randomly chosen fluorescent light bulbs, a specific number will be defective. For simplicity, let's denote the probability of a bulb being defective as ( p ) and the probability of a bulb not being defective as ( q ).

Defining Probabilities

Given that 10% of the bulbs are defective, we can write:

( p 0.1 )

( q 1 - p 0.9 )

Calculating Probabilities with Binomial Distribution

To find the probability of having a specific number of defective bulbs in a sample of 6, we can use the binomial distribution formula:

( P(X k) binom{n}{k} p^k q^{n-k} )

Where:

( n ) is the number of trials (in this case, 6) ( k ) is the number of successes (defective bulbs) ( binom{n}{k} ) is the binomial coefficient, calculated as ( frac{n!}{k!(n-k)!} ) ( p ) is the probability of success on a single trial ( q ) is the probability of failure (non-defective) on a single trial

We need to calculate for ( k 0, 1, 2, 3, 4, 5, 6 ).

Calculation for Each Scenario

No Defective Bulbs (k0)

P(0 defective bulbs) ( binom{6}{0} (0.1)^0 (0.9)^6 )

( 1 times 1 times 0.531441 0.531441 )

One Defective Bulb (k1)

P(1 defective bulb) ( binom{6}{1} (0.1)^1 (0.9)^5 )

( 6 times 0.1 times 0.59049 0.354294 )

Two Defective Bulbs (k2)

P(2 defective bulbs) ( binom{6}{2} (0.1)^2 (0.9)^4 )

( 15 times 0.01 times 0.6561 0.098415 )

Three Defective Bulbs (k3)

P(3 defective bulbs) ( binom{6}{3} (0.1)^3 (0.9)^3 )

( 20 times 0.001 times 0.729 0.01458 )

Four Defective Bulbs (k4)

P(4 defective bulbs) ( binom{6}{4} (0.1)^4 (0.9)^2 )

( 15 times 0.0001 times 0.81 0.001215 )

Five Defective Bulbs (k5)

P(5 defective bulbs) ( binom{6}{5} (0.1)^5 (0.9)^1 )

( 6 times 0.00001 times 0.9 0.000054 )

Six Defective Bulbs (k6)

P(6 defective bulbs) ( binom{6}{6} (0.1)^6 (0.9)^0 )

( 1 times 0.000001 times 1 0.000001 )

Conclusion and Assumptions

The calculation of these probabilities assumes that each bulb is independent and the probability of failure remains constant. If the sample size is large relative to the population, the assumption that the trial outcomes are independent can be valid. However, if the sample size is significant compared to the population, the outcomes might not be independent.

Overall, the probabilities for the different scenarios are:

( P(text{0 defective}) 0.531441 )

( P(text{1 defective}) 0.354294 )

( P(text{2 defective}) 0.098415 )

( P(text{3 defective}) 0.01458 )

( P(text{4 defective}) 0.001215 )

( P(text{5 defective}) 0.000054 )

( P(text{6 defective}) 0.000001 )

The importance of these probabilities is that they help the factory understand the likelihood of finding a certain number of defective bulbs, which can inform quality control measures and ensure product reliability.