Probability of Selecting Non-Defective Bulbs and Candles from a Box

Probability is a fundamental concept in mathematics and statistics that helps us understand the likelihood of events occurring. This article explores the probability of selecting non-defective items from a box of bulbs and candles using combinatorial methods. By understanding these principles, one can apply them to various real-world scenarios, from manufacturing defects to quality control.

Introduction to Probability and Combinatorics

Probability is defined as the number of favorable outcomes over the total number of outcomes. Combinatorics, a branch of mathematics, deals with counting objects in finite sets. These concepts are essential for solving problems involving random selection and determining outcomes with specific conditions.

Problem Statement

We are given a box of 10 bulbs, out of which 4 are defective. If 3 bulbs are chosen at random, what is the probability that none of them is defective?

Solution using Combinatorial Methods

Step 1: Determine the Total Number of Bulbs and Defective Bulbs

Total bulbs 10

Defective bulbs 4

Non-defective bulbs 10 - 4 6

Step 2: Calculate the Total Ways to Choose 3 Bulbs from 10

The total number of ways to choose 3 bulbs from 10 is given by the combination formula:

$$binom{n}{r} frac{n!}{r! (n-r)!}$$

So, the total ways to choose 3 bulbs from 10 is:

$$binom{10}{3} frac{10!}{3! (10-3)!} frac{10 times 9 times 8}{3 times 2 times 1} 120$$

Step 3: Calculate the Ways to Choose 3 Non-Defective Bulbs from 6

The number of ways to choose 3 non-defective bulbs from 6 is:

$$binom{6}{3} frac{6!}{3! (6-3)!} frac{6 times 5 times 4}{3 times 2 times 1} 20$$

Step 4: Calculate the Probability that None of the Chosen Bulbs are Defective

The probability (P) that none of the chosen bulbs are defective is the ratio of the number of ways to choose 3 non-defective bulbs to the total number of ways to choose 3 bulbs:

$$P frac{binom{6}{3}}{binom{10}{3}} frac{20}{120} frac{1}{6}$$

Conclusion

The probability that none of the 3 bulbs chosen are defective is (frac{1}{6}).

Probability of Selecting Non-Defective Candles

Now, let's calculate the probability that 3 out of 6 candles are non-defective, given that the probability of a candle being defective is (frac{4}{10}0.4).

Step 1: Determine the Total Number of Outcomes

The total number of ways to choose 6 candles from 10 is:

$$binom{10}{6} frac{10!}{6! (10-6)!} frac{10 times 9 times 8 times 7}{4 times 3 times 2 times 1} 210$$

Step 2: Determine the Number of Favourable Outcomes

The number of ways to choose 3 defective candles from 4 is:

$$binom{4}{3} frac{4!}{3! (4-3)!} frac{4 times 3 times 2}{3 times 2 times 1} 4$$

The number of ways to choose 3 non-defective candles from 6 is:

$$binom{6}{3} frac{6!}{3! (6-3)!} frac{6 times 5 times 4}{3 times 2 times 1} 20$$

Therefore, the total number of ways to select 3 defective candles and 3 non-defective candles is:

$$4 times 20 80$$

Step 3: Calculate the Probability

Thus, the probability of drawing 3 defective candles from a selection of 6 is:

$$P frac{80}{210} frac{8}{21} approx 0.38$$

Verification using the Probability Space

To verify, we can calculate all instances of drawing defective candles and total outcomes to ensure accuracy.

- Drawing 0 defective candles and 6 non-defective candles: 1 way - Drawing 1 defective candle and 5 non-defective candles: 24 ways - Drawing 2 defective candles and 4 non-defective candles: 90 ways - Drawing 3 defective candles and 3 non-defective candles: 80 ways - Drawing 4 defective candles and 2 non-defective candles: 15 ways - Drawing 5 defective candles and 1 non-defective candle: 0 ways - Drawing 6 defective candles and 0 non-defective candles: 0 ways

Total number of outcomes 1 24 90 80 15 0 0 210

Conclusion

This problem emphasizes the application of probability and combinatorics in real-world scenarios. Understanding these methods helps in making informed decisions in various fields such as quality control, manufacturing, and data analysis.