Bernoulli Trials

Bernoulli Trials are a fundamental concept in probability theory and statistics, forming the basis for understanding more complex distributions such as the binomial distribution. Named after the Swiss mathematician Jacob Bernoulli, Bernoulli trials refer to a series of independent experiments, each having exactly two possible outcomes: success and failure. The study of Bernoulli trials not only lays the groundwork for theoretical probability but also has practical applications in a wide range of fields, from finance and engineering to biology and sports analytics.

1.0What are Bernoulli Trials?

A Bernoulli trial is a random experiment where:

1. There are exactly two possible outcomes: These outcomes are often referred to as success and failure. Success and failure are subjective and depend on the context of the problem. For example, when tossing a coin, getting a head can be considered a success, while a tail would be a failure.
2. Each trial is independent: The result of one trial has no impact on the result of any other trial. This independence is crucial for maintaining consistency in probability calculations.
3. The probability of success stays constant: The probability of success, denoted as p, remains unchanged throughout all the trials. Similarly, the probability of failure, q, is given by q = 1 – p.

A typical example of a Bernoulli trial is flipping a fair coin. Each flip yields one of two outcomes—heads or tails—and the probability of getting heads (considered a success) remains fixed at 0.5 for every single flip.

2.0Bernoulli Trials and Binomial Distribution

When a series of Bernoulli trials is conducted, we are often interested in the number of successes that occur over a fixed number of trials. This is where the binomial distribution comes into play. The binomial distribution describes the probability of having exactly k successes in n independent Bernoulli trials, each with a success probability p.

The probability mass function (PMF)associated with a binomial random variable X, which tallies the number of successes, is given by:

$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$

where:

• $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is the binomial coefficient, which indicates the number of ways to select k successes out of n trials,
• p represents the probability of success on a single trial,
• (1 – p) is the probability of failure on a single trial,
• n is the total number of trials, and
• k is the number of successes.

This formula shows that the binomial distribution is directly derived from Bernoulli trials, making the two concepts closely intertwined. The expected value (mean) of a binomial random variable X is expressed as E(X) = np, and the variance is Var(X) = np (1 – p).

3.0Applications of Bernoulli Trials and Binomial Distribution

Bernoulli trials and the associated binomial distribution have numerous applications:

• Quality Control: Used to determine the probability of a certain number of defective products in a batch.
• Medical Studies: Helps in understanding the likelihood of a certain number of patients responding positively to a new treatment.
• Sports Analytics: Determines the probability of a team winning a certain number of games in a series.
• Finance: Models the likelihood of success in a sequence of investments or trades.

4.0Solved Example of Bernoulli Trials

Example 1: Suppose we flip a fair coin 10 times. What is the probability/chance of obtaining exactly 6 heads?

Solution:

Here, each flip is a Bernoulli trial with p = 0.5. The problem asks for the probability of getting 6 heads (successes) out of 10 flips (trials). We can use the binomial distribution formula:

$P(X=6)=\left(\genfrac{}{}{0ex}{}{10}{6}\right)(0.5{)}^6(0.5{)}^4$

$\text{ Calculating }\left(\genfrac{}{}{0ex}{}{10}{6}\right)=210\text{, we get: }$

$P(X=6)=210\times (0.5{)}^{10}=210\times \frac{1}{1024}\approx 0.205$

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

Example 2: In a factory, it is known that 2% of the produced items are defective. If we randomly select 50 items, what is the probability that exactly 2 of them are defective?

Solution:

Here, p = 0.02, n = 50, and k = 2. Using the binomial distribution formula:

$P(X=2)=\left(\genfrac{}{}{0ex}{}{50}{2}\right)(0.02{)}^2(0.98{)}^{48}$

$\text{ Calculating }\left(\genfrac{}{}{0ex}{}{50}{2}\right)=1225\text{, we get }$

$P(X=2)=1225\times (0.02{)}^2\times (0.98{)}^{48}\approx 0.294$

Thus, the probability of finding exactly 2 defective items in a sample of 50 is approximately 0.294 or 29.4%.

Example 3: Suppose you have a fair coin, and you flip it 5 times. What is the probability of getting exactly 3 heads in these 5 flips?

Solution:

1. Identifying Parameters:  

• Each flip of the coin is a Bernoulli trial because there are only two possible outcomes: heads (success) or tails (failure).  
• The probability of getting heads (success) in a single trial is p = 0.5.  
• The probability of getting tails (failure) is q = 1 – p = 0.5.  
• The number of trials is n = 5.  
• The number of successes we are interested in is k = 3.

2. Using the Binomial Distribution Formula: 

For n independent Bernoulli trials, the probability of getting exactly k successes is given by the binomial distribution formula:

$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$

Substituting the values:  

$P(X=3)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)(0.5{)}^3(0.5{)}^{5-3}$

3. Calculating the Binomial Coefficient:  

The binomial coefficient $\left(\genfrac{}{}{0ex}{}{5}{3}\right)$ represents the number of ways to choose 3 successes out of 5 trials and is calculated as:

$\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!(5-3)!}=\frac{5\times 4\times 3!}{3!\times 2\times 1}=10$

4. Substitute the Binomial Coefficient and Probabilities:

$P(X=3)=10\times (0.5{)}^3\times (0.5{)}^2$

$P(X=3)=10\times (0.5{)}^5$

5. Simplifying:

$P(X=3)=10\times 0.03125=0.3125$

Final Answer: The probability of getting exactly 3 heads in 5 flips of a fair coin is 0.3125 or 31.25%.

5.0Practice Questions on Bernoulli Trials

1. A coin is tossed 10 times. What is the probability of getting exactly 7 heads?
2. A batch of products has a 4% defect rate. If a quality inspector randomly selects 25 items, what is the probability that exactly 3 of them are defective?
3. A student takes a 15-question multiple-choice quiz where each question has 4 options. If the student guesses each answer randomly, what is the probability that they get exactly 5 questions correct?
4. In a retail store, there is a 20% chance that a customer will make a purchase. If 50 customers visit the store in a day, what is the probability that exactly 10 customers make a purchase?
5. A basketball player has an 85% probability of making a free throw. If they take 20 free throws in a game, what is the probability that they make exactly 18 successful shots?
6. A factory produces light bulbs, and 2% of them are found to be defective. If a quality control inspector checks 100 bulbs, what is the probability that exactly 5 of them are defective?

6.0Sample Questions on Bernoulli Trials

1. What is the formula for calculating the probability of success in Bernoulli Trials?  

Ans: The probability of getting exactly k successes in n Bernoulli trials is given by the formula:

$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$

where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is the binomial coefficient, p is the probability of success, and (1 – p) is the probability of failure.

2. What does the binomial coefficient represent?  

Ans: The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ represents the number of ways to choose k successes from n trials. It is calculated as:

$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$