Faulty Light Bulbs Probability

Solution & Explanation

The problem at hand involves calculating the probability of exactly 2 defective light bulbs out of a box of 10, given that each bulb has a 5% chance of being defective. This scenario is modeled by a binomial distribution, which is used to determine the probability of a fixed number of successes in a fixed number of independent Bernoulli trials.

Step-by-Step Breakdown:

1. Understanding the Binomial Distribution

   • n: Total number of trials (light bulbs) = 10
   • k: Number of successful outcomes (faulty bulbs) = 2
   • p: Probability of success on an individual trial (faulty bulb) = 0.05
   • q: Probability of failure on an individual trial (non-faulty bulb) = 1 - p = 0.95

2. Binomial Probability Formula

   The probability of getting exactly k successes in n trials is given by:

   $P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$

   Where $\binom{n}{k}$ is the binomial coefficient, calculated as:

   $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

3. Calculate the Binomial Coefficient

   $\binom{10}{2} = \frac{10!}{2! \cdot (10-2)!} = \frac{10 \cdot 9}{2 \cdot 1} = 45$

4. Calculate the Probability

   Plugging the values into the formula:

   $P(X = 2) = 45 \cdot (0.05)^2 \cdot (0.95)^8$

   • Calculate $(0.05)^2$: $(0.05)^2 = 0.0025$

   • Calculate $(0.95)^8$: $(0.95)^8 \approx 0.6634$

   • Combine the results: $P(X = 2) = 45 \cdot 0.0025 \cdot 0.6634$ $P(X = 2) \approx 0.0746$

5. Interpretation

   The probability that exactly 2 out of the 10 bulbs in the package are defective is approximately 0.0746, or 7.46%.

This result indicates that there is a 7.46% chance of finding exactly two defective bulbs in a randomly selected box of 10 bulbs, given the defect rate of 5% per bulb.