The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials of a binary experiment. Each trial has two possible outcomes: success or failure. The binomial distribution is characterized by two parameters: the number of trials \( n \) and the probability of success \( p \).

1. Binary Experiment: An experiment with two possible outcomes, often labeled as "success" and "failure".
2. Number of Trials (n): The fixed number of times the binary experiment is conducted.
3. Probability of Success (p): The probability that a single trial results in success.
4. Probability of Failure (q): The probability that a single trial results in failure, where \( q = 1 - p \).

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

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

Where:

• \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)
• \( p \) is the probability of success
• \( (1-p) \) is the probability of failure
• \( k \) is the number of successes

Suppose you are flipping a fair coin (where the probability of heads, \( p \), is 0.5) 10 times. What is the probability of getting exactly 6 heads?

Given:

• \( n = 10 \)
• \( p = 0.5 \)
• \( k = 6 \)

Using the binomial probability formula:

\[ P(X = 6) = \binom{10}{6} (0.5)^6 (0.5)^{4} \]

First, calculate the binomial coefficient:

\[ \binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = 210 \]

Then, calculate the probability:

\[ P(X = 6) = 210 \times (0.5)^6 \times (0.5)^4 \] \[ P(X = 6) = 210 \times (0.015625) \times (0.0625) \] \[ P(X = 6) = 210 \times 0.0009765625 \] \[ P(X = 6) \approx 0.205 \]

So, the probability of getting exactly 6 heads in 10 flips of a fair coin is approximately 0.205.

A factory produces light bulbs, and the probability that a bulb is defective is 0.02. If a quality control inspector randomly selects 20 bulbs, what is the probability that exactly 2 bulbs are defective?

Given:

• \( n = 20 \)
• \( p = 0.02 \)
• \( k = 2 \)

Using the binomial probability formula:

\[ P(X = 2) = \binom{20}{2} (0.02)^2 (0.98)^{18} \]

First, calculate the binomial coefficient:

\[ \binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20!}{2!18!} = 190 \]

Then, calculate the probability:

\[ P(X = 2) = 190 \times (0.02)^2 \times (0.98)^{18} \] \[ P(X = 2) = 190 \times 0.0004 \times 0.698337 \] \[ P(X = 2) \approx 0.053 \]

So, the probability that exactly 2 out of 20 bulbs are defective is approximately 0.053.

A basketball player has a free throw success rate of 70%. If she takes 15 free throws in a game, what is the probability that she makes exactly 10 of them?

Given:

• \( n = 15 \)
• \( p = 0.7 \)
• \( k = 10 \)

Using the binomial probability formula:

\[ P(X = 10) = \binom{15}{10} (0.7)^{10} (0.3)^{5} \]

First, calculate the binomial coefficient:

\[ \binom{15}{10} = \frac{15!}{10!(15-10)!} = \frac{15!}{10!5!} = 3003 \]

Then, calculate the probability:

\[ P(X = 10) = 3003 \times (0.7)^{10} \times (0.3)^{5} \] \[ P(X = 10) = 3003 \times 0.0282475249 \times 0.00243 \] \[ P(X = 10) \approx 0.201 \]

So, the probability that she makes exactly 10 out of 15 free throws is approximately 0.201.

1. Incorrect Calculation of Binomial Coefficient: Ensure you correctly compute the binomial coefficient \( \binom{n}{k} \). Use a calculator or software to avoid manual errors.
2. Misinterpretation of \( p \) and \( q \): Remember that \( q = 1 - p \). Double-check your values for \( p \) and \( q \) to ensure accuracy.
3. Rounding Errors: Be cautious with rounding intermediate steps. It's often best to round only the final result to avoid significant errors.

The binomial distribution is a fundamental concept in statistics, useful for modeling scenarios with fixed numbers of independent trials and binary outcomes. Understanding how to calculate binomial probabilities allows you to analyze and interpret various real-world situations effectively. In the next section, we will explore the normal distribution, another critical probability distribution in statistics.