Binomial Probability Distribution Examples and Solutions
The binomial probability distribution is a fundamental concept in statistics that models the number of successes in a fixed number of independent trials, each with the same probability of success. This discrete distribution appears frequently in real-world scenarios such as quality control, medical trials, market research, and sports analytics. Understanding binomial probability through practical examples and step-by-step solutions is essential for students and professionals dealing with data analysis.
What is Binomial Probability Distribution?
Before diving into examples, let's establish the foundation. A binomial distribution applies when an experiment satisfies four conditions:
- Fixed number of trials (n): The experiment consists of a predetermined number of repetitions
- Independent trials: Each trial doesn't affect the outcome of others
- Two possible outcomes: Success or failure for each trial
- Constant probability (p): The probability of success remains the same across all trials
The binomial probability formula is:
P(X = k) = C(n,k) × p^k × (1-p)^(n-k)
Where:
- P(X = k) = probability of exactly k successes
- C(n,k) = number of combinations of n items taken k at a time
- p = probability of success on a single trial
- n = total number of trials
- k = number of successes
Example 1: Quality Control in Manufacturing
A factory produces light bulbs, and historical data shows that 5% of bulbs are defective. A quality inspector randomly selects 20 bulbs from a shipment for testing.
Question: What is the probability that exactly 2 bulbs are defective?
Solution:
Given:
- n = 20 (total bulbs tested)
- k = 2 (defective bulbs we're interested in)
- p = 0.05 (probability of a bulb being defective)
Using the binomial formula: P(X = 2) = C(20,2) × (0.05)^2 × (0.95)^18
First, calculate C(20,2): C(20,2) = 20!/(2!(20-2)!) = 190
Now, substitute: P(X = 2) = 190 × 0.Here's the thing — 0025 × 0. 3774 P(X = 2) = 0.
Answer: There's approximately a 17.89% chance that exactly 2 out of 20 bulbs are defective.
Example 2: Medical Treatment Effectiveness
A new vaccine has a 70% success rate in preventing a particular disease. If 15 patients receive the vaccine, what is the probability that at least 10 will be protected?
Solution:
For "at least 10," we need P(X ≥ 10) = P(10) + P(11) + P(12) + P(13) + P(14) + P(15)
Given:
- n = 15
- p = 0.70
- We need the sum of probabilities from k = 10 to 15
Calculating each term:
P(X = 10) = C(15,10) × (0.Plus, 70)^10 × (0. 30)^5 = 3003 × 0.Day to day, 0282 × 0. 00243 = 0.
P(X = 11) = C(15,11) × (0.70)^11 × (0.On the flip side, 30)^4 = 1365 × 0. Still, 0197 × 0. 0081 = 0.
P(X = 12) = C(15,12) × (0.70)^12 × (0.Practically speaking, 30)^3 = 455 × 0. 0138 × 0.027 = 0.
P(X = 13) = C(15,13) × (0.In real terms, 70)^13 × (0. 0097 × 0.30)^2 = 105 × 0.09 = 0.
P(X = 14) = C(15,14) × (0.70)^14 × (0.0068 × 0.30)^1 = 15 × 0.30 = 0 No workaround needed..
P(X = 15) = C(15,15) × (0.30)^0 = 1 × 0.But 70)^15 × (0. 0047 × 1 = 0 The details matter here..
Summing all probabilities: P(X ≥ 10) = 0.That's why 2061 + 0. 0916 + 0.Plus, 1700 + 0. 2186 + 0.0306 + 0.0047 = 0 Practical, not theoretical..
Answer: There's approximately a 72.16% probability that at least 10 out of 15 patients will be protected by the vaccine.
Example 3: Multiple Choice Test Performance
A student takes a multiple-choice test with 20 questions, each having 4 options (only one correct). If the student guesses randomly on all questions, what is the probability that they:
a) Answer exactly 5 questions correctly? b) Answer no more than 3 questions correctly?
Solution:
For random guessing, p = 1/4 = 0.25
Part a: P(X = 5)
P(X = 5) = C(20,5) × (0.Because of that, 75)^15 C(20,5) = 15504 P(X = 5) = 15504 × 0. 25)^5 × (0.Also, 0009766 × 0. 01336 P(X = 5) = 0 Simple, but easy to overlook..
Part b: P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)
P(X = 0) = C(20,0) × (0.25)^0 × (0.75)^20 = 1 × 1 × 0.00317 = 0.
P(X = 1) = C(20,1) × (0.Still, 25)^1 × (0. 75)^19 = 20 × 0.25 × 0.00423 = 0.
P(X = 2) = C(20,2) × (0.75)^18 = 190 × 0.And 25)^2 × (0. 0625 × 0.00563 = 0.
P(X = 3) = C(20,3) × (0.But 25)^3 × (0. 75)^17 = 1140 × 0.015625 × 0.00751 = 0.
P(X ≤ 3) = 0.0669 + 0.0212 + 0.That's why 00317 + 0. 1339 = 0 Most people skip this — try not to..
Answers: a) Probability of exactly 5 correct answers: 20.23% b) Probability of no more than 3 correct answers: 22.52%
Example 4: Sports Analytics
A basketball player makes free throws with a 75% success rate. In a game where they attempt 12 free throws, what is the probability that they make at least 8?
Solution:
We need P(X ≥ 8) = P(8) + P(9
