Understanding the Probability of Selecting at Least One Defective Light Bulb in a Box with Replacement

In this article, we will explore the probability of selecting at least one defective light bulb from a box containing 12 light bulbs, out of which 5 are defective, using the method of complementary probability and the fundamental principles of probability theory. This analysis is crucial for various applications, including quality control and reliability engineering.

Introduction to the Problem

Consider a box containing 12 light bulbs, of which 5 are defective. All bulbs are identical and have an equal probability of being chosen. We will select two light bulbs with replacement, meaning the first bulb is put back into the box before the second selection. The goal is to find the probability that at least one of the selected bulbs is defective.

Step-by-Step Analysis

To tackle this problem, we will use the complementary probability approach. This method simplifies the calculation by focusing on the probability of the opposite event (in this case, selecting no defective bulbs) and then subtracting it from 1.

Step 1: Calculate the Probability of Choosing a Non-Defective Bulb

There are 7 non-defective bulbs (12 total bulbs - 5 defective bulbs). The probability of selecting a non-defective bulb in one draw is:

[ P(text{non-defective}) frac{7}{12} ]

Step 2: Calculate the Probability of Selecting Two Non-Defective Bulbs with Replacement

Since the bulbs are selected with replacement, the draws are independent. Therefore, the probability of selecting two non-defective bulbs in succession is:

[ P(text{both non-defective}) P(text{non-defective}) times P(text{non-defective}) left(frac{7}{12}right) times left(frac{7}{12}right) left(frac{7}{12}right)^2 frac{49}{144} ]

Step 3: Calculate the Probability of Selecting at Least One Defective Bulb

The probability of selecting at least one defective bulb can be found by subtracting the probability of selecting both non-defective bulbs from 1:

[ P(text{at least one defective}) 1 - P(text{both non-defective}) 1 - frac{49}{144} frac{144 - 49}{144} frac{95}{144} ]

Thus, the probability that at least one of the selected bulbs is defective is:

[ boxed{frac{95}{144}} ]

Alternative Approach

Another way to look at the problem is to calculate the probability that neither of the selected bulbs is defective. This can be done by directly calculating the product of the probabilities of selecting a non-defective bulb in each draw: [ P(text{neither defective}) frac{7}{12} times frac{7}{12} frac{49}{144} ] The probability that at least one bulb is defective then becomes: [ P(text{at least one defective}) 1 - frac{49}{144} frac{95}{144} ]

Complementary Probability and Replacement Selection

The complementary probability approach is particularly useful when dealing with problems involving replacement. Here, we have used it to avoid working with the complex combinatorial calculations needed to consider all possible outcomes where at least one bulb is defective.

By focusing on the probability of the opposite (non-defective bulbs), we can simplify the problem significantly. This method is widely used in probability and statistics for its clarity and computational efficiency.

Conclusion

In conclusion, the probability of selecting at least one defective bulb from a box of 12 light bulbs, where 5 are defective, when two bulbs are selected with replacement, is (frac{95}{144}). This result is obtained using the complementary probability approach, which simplifies the calculation by focusing on the probability of the complementary event (selecting no defective bulbs) and then subtracting it from 1.

Understanding and applying these principles are valuable for various fields, including quality control, reliability engineering, and data science. By mastering such techniques, one can make informed decisions and optimize processes for better outcomes.