Binomial Distribution Examples With Solutions

Understanding and Applying the Binomial Distribution: Examples with Solutions

The binomial distribution is a fundamental concept in probability and statistics. It describes the probability of getting exactly k successes in n independent Bernoulli trials, where each trial has only two possible outcomes (success or failure), and the probability of success remains constant throughout the trials. Understanding the binomial distribution is crucial for analyzing various real-world scenarios, from quality control in manufacturing to medical research. This article will delve into the binomial distribution, providing clear explanations, worked-out examples, and solutions to help you grasp this vital statistical tool.

What is a Binomial Distribution?

A binomial distribution is characterized by four key parameters:

• n: The number of trials. This is a fixed, positive integer.
• k: The number of successes. This can range from 0 to n.
• p: The probability of success in a single trial. This is a constant value between 0 and 1.
• q: The probability of failure in a single trial. This is equal to 1 - p.

The probability of obtaining exactly k successes in n trials is given by the binomial probability formula:

P(X = k) = (nCk) * p^k * q^(n-k)

where (nCk) represents the binomial coefficient, also written as "n choose k," and calculated as:

(nCk) = n! / (k! * (n-k)!)

where "!" denotes the factorial (e.g., 5! = 54321).

Applying the Binomial Distribution: Examples and Solutions

Let's explore several examples to solidify your understanding of the binomial distribution and its application.

Example 1: Coin Tosses

Suppose you toss a fair coin 10 times. What is the probability of getting exactly 6 heads?

• n = 10 (number of tosses)
• k = 6 (number of heads)
• p = 0.5 (probability of getting a head in a single toss)
• q = 0.5 (probability of getting a tail in a single toss)

Using the binomial probability formula:

P(X = 6) = (10C6) * (0.5)⁶ * (0.5)^(10-6)

(10C6) = 10! / (6! * 4!) = 210

P(X = 6) = 210 * (0.5)⁶ * (0.5)⁴ = 210 * 0.015625 * 0.0625 ≈ 0.205

Therefore, the probability of getting exactly 6 heads in 10 tosses is approximately 0.205 or 20.5%.

Example 2: Defective Products

A factory produces light bulbs. The probability that a light bulb is defective is 0.02. If you randomly select 20 light bulbs, what is the probability that exactly 2 are defective?

• n = 20 (number of light bulbs selected)
• k = 2 (number of defective light bulbs)
• p = 0.02 (probability of a single light bulb being defective)
• q = 0.98 (probability of a single light bulb being non-defective)

P(X = 2) = (20C2) * (0.02)² * (0.98)¹⁸

(20C2) = 20! / (2! * 18!) = 190

P(X = 2) = 190 * (0.0004) * (0.694) ≈ 0.0528

The probability that exactly 2 out of 20 light bulbs are defective is approximately 0.0528 or 5.28%.

Example 3: Multiple Choice Test

A multiple-choice test has 15 questions, each with 4 options. A student guesses randomly on each question. What's the probability the student gets exactly 5 questions correct?

• n = 15 (number of questions)
• k = 5 (number of correct answers)
• p = 0.25 (probability of answering a single question correctly by guessing)
• q = 0.75 (probability of answering a single question incorrectly by guessing)

P(X = 5) = (15C5) * (0.25)⁵ * (0.75)¹⁰

(15C5) = 3003

P(X = 5) = 3003 * (0.0009765625) * (0.0563135147) ≈ 0.1651

The probability of getting exactly 5 questions correct by random guessing is approximately 0.1651 or 16.51%.

Example 4: Medical Trials

A new drug is being tested. In a trial with 50 participants, the probability that a participant experiences a side effect is 0.1. What is the probability that exactly 3 participants will experience a side effect?

• n = 50 (number of participants)
• k = 3 (number of participants experiencing side effects)
• p = 0.1 (probability of a participant experiencing a side effect)
• q = 0.9 (probability of a participant not experiencing a side effect)

P(X=3) = (50C3) * (0.1)³ * (0.9)⁴⁷

(50C3) = 19600

Calculating this directly can be cumbersome. For larger values of 'n', using statistical software or calculators is recommended. Many calculators and software packages have built-in functions to compute binomial probabilities directly.

Understanding the Binomial Distribution's Shape

The shape of the binomial distribution depends on the values of n and p.

• Symmetrical: When p = 0.5, the distribution is symmetrical around the mean (n*p).
• Skewed Right: When p < 0.5, the distribution is skewed to the right (positively skewed). The tail on the right extends further than the tail on the left.
• Skewed Left: When p > 0.5, the distribution is skewed to the left (negatively skewed). The tail on the left extends further than the tail on the right.

As n increases, the binomial distribution becomes more symmetrical, regardless of the value of p, approaching a normal distribution. This is described by the Central Limit Theorem.

Mean and Variance of the Binomial Distribution

The mean (expected value) and variance of a binomial distribution are easily calculated:

• Mean (μ): μ = n * p
• Variance (σ²): σ² = n * p * q
• Standard Deviation (σ): σ = √(n * p * q)

These parameters provide valuable insights into the distribution's central tendency and spread.

Cumulative Binomial Probability

Often, we're interested in the probability of getting at least or at most a certain number of successes, rather than exactly a specific number. This requires calculating the cumulative binomial probability. This involves summing the probabilities for all values of k within the desired range.

For example, the probability of getting at least 3 successes in 10 trials (with p = 0.2) would be:

P(X ≥ 3) = P(X = 3) + P(X = 4) + ... + P(X = 10)

Again, statistical software or calculators are highly useful for efficiently computing cumulative binomial probabilities.

Frequently Asked Questions (FAQ)

Q1: What are the assumptions of the binomial distribution?

The binomial distribution relies on four key assumptions:

1. Fixed number of trials (n): The number of trials is predetermined and does not change.
2. Independent trials: The outcome of one trial does not affect the outcome of any other trial.
3. Two outcomes: Each trial results in one of two mutually exclusive outcomes (success or failure).
4. Constant probability of success (p): The probability of success remains constant for all trials.

Q2: When should I not use the binomial distribution?

The binomial distribution is not appropriate when the assumptions above are violated. For example, if the probability of success changes from trial to trial, or if the trials are dependent, you'll need a different probability model.

Q3: How can I calculate binomial probabilities quickly?

For larger values of n, manual calculations become tedious. Statistical software packages (like R, SPSS, or Python's SciPy library) and many scientific calculators offer built-in functions for calculating binomial probabilities and cumulative probabilities efficiently. Online binomial calculators are also readily available.

Conclusion

The binomial distribution is a powerful tool for modeling and analyzing events with binary outcomes. By understanding its formula, assumptions, and applications, you can effectively solve a wide range of probability problems across various fields. Remember to carefully consider the underlying assumptions before applying the binomial distribution to ensure the accuracy and reliability of your results. Mastering the binomial distribution is a cornerstone of statistical literacy, opening doors to more advanced statistical concepts and analyses. Remember to practice with various examples and utilize the available computational tools to make the process efficient and less prone to errors. The examples provided here should serve as a solid foundation for further exploration and application of this crucial statistical concept.