Binomial Probability Calculation
Calculate the probability of a specific number of successes using the binomial distribution formula in a quality control context.
Problem Scenario
A factory produces electronic components with a known defect rate of 8%. A quality inspector randomly selects 12 components from a large production batch. What is the probability that exactly 2 of the 12 components are defective?
Given Data
Requirements
- Verify the conditions for a binomial distribution
- Apply the binomial probability formula to find P(X = 2)
- Interpret the result in context
Solution
Step 1:
Verify binomial conditions: (1) Fixed number of trials: n = 12. (2) Each trial has two outcomes: defective or not defective. (3) Probability of defect is constant: p = 0.08. (4) Trials are independent (assuming the batch is large). All conditions are met.
Step 2:
Write the binomial formula: P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where C(n, k) is the number of combinations.
Step 3:
Calculate the combination: C(12, 2) = 12! / (2! * 10!) = (12 * 11) / (2 * 1) = 66.
Step 4:
Substitute values: P(X = 2) = 66 * (0.08)^2 * (0.92)^10 = 66 * 0.0064 * 0.4344 = 66 * 0.002780 = 0.1835.
Step 5:
Interpretation: There is approximately an 18.35% probability that exactly 2 out of 12 randomly selected components will be defective.
Final Answer
P(X = 2) = 0.1835 or approximately 18.35%. There is about an 18.4% chance that exactly 2 of the 12 selected components are defective.
Key Takeaways
- ✓The binomial distribution requires fixed n, independent trials, constant probability of success, and exactly two outcomes per trial.
- ✓The mean of a binomial distribution is mu = np and the standard deviation is sigma = sqrt(np(1-p)). Here, mu = 12(0.08) = 0.96 defects on average.
- ✓For probabilities like P(X >= 2) or P(X <= 2), you can sum individual binomial probabilities or use the complement rule: P(X >= 2) = 1 - P(X = 0) - P(X = 1).
Common Errors to Avoid
- ✗Confusing the combination formula with the permutation formula. Combinations C(n,k) = n! / [k!(n-k)!] do not consider order, while permutations P(n,k) = n! / (n-k)! do.
- ✗Applying the binomial model when trials are not independent or when the probability changes from trial to trial. In those cases, the hypergeometric or other distributions may be appropriate.
- ✗Forgetting to raise (1-p) to the power (n-k) rather than to n. The exponent on q must equal the number of failures, not the total trials.
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Common questions about this problem type
Use the exact binomial formula when n is small or when you need precise probabilities. The normal approximation to the binomial is appropriate when both np >= 10 and n(1-p) >= 10. In this problem, np = 0.96 < 10, so the normal approximation would not be appropriate.
Use the complement rule: P(X >= 2) = 1 - P(X = 0) - P(X = 1). Calculate P(X = 0) = C(12,0)(0.08)^0(0.92)^12 = 0.3677, and P(X = 1) = C(12,1)(0.08)^1(0.92)^11 = 0.3837. Then P(X >= 2) = 1 - 0.3677 - 0.3837 = 0.2486.