Calculating the Probability of Selecting Exactly One Defective Ball from a Box

In a statistical probability problem, we might be interested in the scenario where selecting a certain number of balls from a mixed box has a particular outcome. For instance, if we have a box containing 15 balls, out of which 5 are defective and 10 are non-defective, what is the probability of picking exactly one defective ball when we select 5 balls at random?

Using Hypergeometric Distribution

To solve this problem, we can use the hypergeometric distribution, which is a probability distribution that describes the number of successes in a fixed number of draws without replacement from a finite population.

Step-by-Step Solution

Understand the Parameters: Total balls, N 15 Total defective balls, K 5 Total non-defective balls 10 Number of balls selected, n 5 Number of defective balls to be selected, k 1 Hypergeometric Probability Formula:

P(Xk) (frac{binom{K}{k} cdot binom{N-K}{n-k}}{binom{N}{n}})

Calculate the Components: Calculate (binom{K}{k}): The number of ways to choose 1 defective ball from 5.

(binom{5}{1} 5)

Calculate (binom{N-K}{n-k}): The number of ways to choose 4 non-defective balls from 10.

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

Calculate (binom{N}{n}): The total number of ways to choose 5 balls from 15.

(binom{15}{5} frac{15!}{5!(15-5)!} frac{15 times 14 times 13 times 12 times 11}{5 times 4 times 3 times 2 times 1} 3003)

Substitute the Values into the Formula:

(P(X1) frac{binom{5}{1} cdot binom{10}{4}}{binom{15}{5}} frac{5 cdot 210}{3003} frac{1050}{3003})

Simplify the Fraction:

To simplify (frac{1050}{3003}), we find the greatest common divisor (GCD) of 1050 and 3003, which is 21. Dividing both the numerator and the denominator by 21, we get (frac{1050 div 21}{3003 div 21} frac{50}{143})

Final Result:

The probability of selecting exactly one defective ball when choosing 5 balls is: (boxed{frac{50}{143}})

Alternative Probability Calculation

Another approach involves using the binomial distribution, especially if we have a large number of components and a small probability of defect, as in the second part of the problem.

Bernoulli Trials and Binomial Distribution

In the second scenario, if 5 of the 15 components are defective and we need to find the probability for 5 out of 100 components to be defective, we can use the binomial distribution with:

Number of trials, n 100 Probability of success, p 0.05 (since 5 out of 15 are defective) Number of successes, k 95 (since 95 out of 100 are non-defective)

Using the binomial probability formula:

(P(X1) 100! / 1! 99! times (0.05) times (0.95)^9)

Calculating this results in approximately 0.031512, or 3.2%.

The distribution of the number of defects can be visualized on a graph with the x-axis representing the number of defects and the y-axis representing their associated probabilities. This graph would show that about 4 out of 100 boxes contain defects.

Conclusion

The hypergeometric and binomial distributions provide useful tools for solving problems involving the probability of specific outcomes in a sampling process. Understanding these distributions can help in making informed decisions based on statistical data.