Probability of Exactly 2 Defective Items in 20
Suppose each item is independently defective with probability 0.08, and a sample contains 20 items. The binomial model answers an exact-count question by combining the number of possible item combinations with the probability of one arrangement.
The question
In a sample of 20 independent items, each with an 8% defect probability, what is the chance that exactly 2 are defective?
These calculations use a probability model that matches the stated assumptions. If the events are not independent or the chance changes between trials, the binomial formula may not describe the real situation.
Step-by-step solution
- Set n = 20 items, k = 2 defective items, and p = 0.08. The probability an item is not defective is 1 − p = 0.92.
- Choose which 2 of the 20 items are defective: C(20, 2) = 190.
- For any chosen pair, the probability of that arrangement is (0.08)² × (0.92)¹⁸.
- Multiply: C(20, 2)(0.08)²(0.92)¹⁸ ≈ 0.271091.
Answer: The probability is about 0.2711, or 27.11%. This assumes each item has the same defect chance and one item’s condition does not affect another’s.
Check the result with a calculator
Open the binomial calculator with n = 20, p = 0.08 and k = 2. You can change the inputs to compare a different number of trials, success probability or target count.
For other probability questions, use the binomial distribution calculator for exact, cumulative and range outcomes, or browse the full calculator directory.