Probability – Bernoulli Trials and Binomial Distribution

Bernoulli Trials

  • A sequence of independent experiments with only two possible outcomes: success ($S$) and failure ($F$).
  • The probability of success in each trial is denoted by $p$, and the probability of failure is $1-p$.
  • Examples: flipping a coin, rolling a die, making a free throw in basketball.

Binomial Distribution

  • Binomial distribution models the number of successes in a fixed number of Bernoulli trials.
  • The number of successes, denoted by $X$, is a discrete random variable.
  • The parameters of the binomial distribution are $n$ (number of trials) and $p$ (probability of success in each trial).

Probability Mass Function (PMF)

  • The probability of obtaining exactly $k$ successes in $n$ trials is given by the binomial probability mass function: $$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$$
  • Here, $\binom{n}{k}$ is the binomial coefficient, given by $\frac{n!}{k!(n-k)!}$, and $0 \leq k \leq n$.

Expected Value and Variance

  • The expected value (mean) of a binomial distribution is given by: $$E(X) = n \cdot p$$
  • The variance of a binomial distribution is given by: $$Var(X) = n \cdot p \cdot (1-p)$$

Example 1

  • A fair coin is tossed 5 times.
  • Find the probability of getting exactly 3 heads. Solution:
  • $n = 5$ (number of trials)
  • $p = 0.5$ (probability of success in each trial)
  • We want to find $P(X = 3)$

Example 1 (continued)

Using the binomial PMF: $$P(X = 3) = \binom{5}{3} \cdot 0.5^3 \cdot (1-0.5)^{5-3}$$ Simplifying:

  • $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \cdot 4}{2 \cdot 1} = 10$
  • $0.5^3 = 0.125$
  • $(1-0.5)^{5-3} = 0.25$

Example 1 (continued)

Substituting the values: $$P(X = 3) = 10 \cdot 0.125 \cdot 0.25 = 0.3125$$ Therefore, the probability of getting exactly 3 heads when tossing a fair coin 5 times is 0.3125.

Example 2

  • A biased coin has a 0.7 probability of landing on heads.
  • If the coin is tossed 10 times, find the probability of getting at most 4 heads. Solution:
  • $n = 10$ (number of trials)
  • $p = 0.7$ (probability of success in each trial)
  • We want to find $P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Example 2 (continued)

Using the binomial PMF:

  • $P(X = 0) = \binom{10}{0} \cdot 0.7^0 \cdot (1-0.7)^{10-0}$
  • $P(X = 1) = \binom{10}{1} \cdot 0.7^1 \cdot (1-0.7)^{10-1}$
  • $P(X = 2) = \binom{10}{2} \cdot 0.7^2 \cdot (1-0.7)^{10-2}$
  • $P(X = 3) = \binom{10}{3} \cdot 0.7^3 \cdot (1-0.7)^{10-3}$
  • $P(X = 4) = \binom{10}{4} \cdot 0.7^4 \cdot (1-0.7)^{10-4}$ Simplifying each term:
  • $\binom{10}{0} = 1$
  • $\binom{10}{1} = 10$
  • $\binom{10}{2} = 45$
  • $\binom{10}{3} = 120$
  • $\binom{10}{4} = 210$

Example 2 (continued)

Finding the probabilities:

  • $P(X = 0) = 1 \cdot 0.7^0 \cdot 0.3^{10}$
  • $P(X = 1) = 10 \cdot 0.7^1 \cdot 0.3^9$
  • $P(X = 2) = 45 \cdot 0.7^2 \cdot 0.3^8$
  • $P(X = 3) = 120 \cdot 0.7^3 \cdot 0.3^7$
  • $P(X = 4) = 210 \cdot 0.7^4 \cdot 0.3^6$

Example 2 (continued)

Calculating each probability:

  • $P(X = 0) = 1 \cdot 1 \cdot 0.000000001 = 0.000000001$
  • $P(X = 1) = 10 \cdot 0.7 \cdot 0.000000000729 = 0.00000000729$
  • $P(X = 2) = 45 \cdot 0.49 \cdot 0.00000005 = 0.00000011$
  • $P(X = 3) = 120 \cdot 0.343 \cdot 0.000003 = 0.00000103$
  • $P(X = 4) = 210 \cdot 0.2401 \cdot 0.000018 = 0.0000882$ Add up the probabilities to get the final result.

Example 2 (continued)

Adding up the probabilities:

  • $P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$
  • $P(X \leq 4) = 0.000000001 + 0.00000000729 + 0.00000011 + 0.00000103 + 0.0000882$ Calculating the sum:
  • $P(X \leq 4) = 0.00008942829$ Therefore, the probability of getting at most 4 heads when tossing a biased coin 10 times is approximately 0.00008942829.

Expected Value and Variance

  • The expected value (mean) of a binomial distribution is given by: $$E(X) = n \cdot p$$
  • The variance of a binomial distribution is given by: $$Var(X) = n \cdot p \cdot (1-p)$$ These formulas allow us to determine the center and spread of a binomial distribution.

Example 3

  • A multiple-choice test has 20 questions.
  • Each question has 4 options, of which only one is correct.
  • A student randomly guesses the answers to all the questions.
  • What is the expected number of correct answers? Solution:
  • $n = 20$ (number of trials/questions)
  • $p = 0.25$ (probability of guessing the correct answer)

Example 3 (continued)

Using the formula for expected value: $$E(X) = n \cdot p$$ Substituting the values: $$E(X) = 20 \cdot 0.25 = 5$$ Therefore, the expected number of correct answers when randomly guessing on a multiple-choice test with 20 questions is 5.

Example 4

  • In a game, a player rolls a fair six-sided die 15 times.
  • The player wins if a 6 appears at least twice.
  • What is the probability of winning? Solution:
  • $n = 15$ (number of trials/rolls)
  • $p = \frac{1}{6}$ (probability of rolling a 6)

Example 4 (continued)

Using the formula for probability:

  • To find the probability of winning, we need to find $P(X \geq 2)$.
  • This can be calculated as the complement of the probability of losing.
  • The probability of losing is the sum of two cases: $P(X = 0)$ and $P(X = 1)$.

Example 4 (continued)

Calculating the probability of losing:

  • $P(X = 0) = \binom{15}{0} \cdot \left(\frac{1}{6}\right)^0 \cdot \left(1-\frac{1}{6}\right)^{15-0}$
  • $P(X = 1) = \binom{15}{1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(1-\frac{1}{6}\right)^{15-1}$ Simplifying each term:
  • $\binom{15}{0} = 1$
  • $\binom{15}{1} = 15$ Calculating the probabilities:
  • $P(X = 0) = 1 \cdot 1 \cdot \left(\frac{5}{6}\right)^{15}$
  • $P(X = 1) = 15 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{14}$ These probabilities will be used to find the probability of winning.

Example 4 (continued)

Calculating the probability of losing:

  • $P(X = 0) = 1 \cdot 1 \cdot \left(\frac{5}{6}\right)^{15}$
  • $P(X = 1) = 15 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{14}$ Simplifying each term:
  • $P(X = 0) = \left(\frac{5}{6}\right)^{15}$
  • $P(X = 1) = \frac{15 \cdot 5^{14}}{6^{15}}$

Example 4 (continued)

Finding the probability of winning:

  • $P(X \geq 2) = 1 - (P(X = 0) + P(X = 1))$ Substituting the values:
  • $P(X \geq 2) = 1 - \left(\left(\frac{5}{6}\right)^{15} + \frac{15 \cdot 5^{14}}{6^{15}}\right)$ Calculating the probability:
  • $P(X \geq 2) \approx 0.3551$ Therefore, the probability of winning in the game is approximately 0.3551.

Properties of Binomial Distribution

  1. The number of trials, $n$, must be a positive integer.
  1. The probability of success, $p$, must be between 0 and 1.
  1. The outcomes of each trial must be independent.
  1. The number of successes, $X$, can take any non-negative integer value from 0 to $n$.
  1. The mean and variance of a binomial distribution can be calculated using the formulas $E(X) = n \cdot p$ and $Var(X) = n \cdot p \cdot (1-p)$.

Applications of Binomial Distribution

  1. Predicting the probability of success or failure in a fixed number of trials.
  1. Analyzing the performance of independent events with two possible outcomes.
  1. Estimating the likelihood of achieving a certain number of successes or failures.
  1. Assessing risk and making decisions based on probabilities.

Limitations of Binomial Distribution

  1. Assumes that the probability of success is constant throughout all trials.
  1. Assumes that the trials are independent of each other.
  1. May not be suitable for situations where the sample size is large or the number of trials is infinite.
  1. Cannot be used for continuous random variables.

Summary

  • Bernoulli trials involve independent experiments with two possible outcomes: success and failure.
  • Binomial distribution models the number of successes in a fixed number of Bernoulli trials.
  • The probability mass function (PMF) of a binomial distribution gives the probability of obtaining a specific number of successes.
  • The expected value (mean) and variance of a binomial distribution can be calculated using formulas.
  • The binomial distribution is commonly used to analyze probabilities and make predictions in various fields.

Key Points to Remember

  • Bernoulli trials have two possible outcomes: success and failure.
  • Binomial distribution models the number of successes in a fixed number of trials.
  • The PMF of a binomial distribution gives the probability of obtaining a specific number of successes.
  • The expected value (mean) and variance of a binomial distribution can be calculated using formulas.
  • Binomial distribution is used for analyzing probabilities and making predictions.

Practice Problems

  1. A fair die is rolled 10 times. Find the probability of getting an odd number at least 3 times.
  1. A student is taking a multiple-choice test with 25 questions. Each question has 5 options. Find the expected number of correct answers if the student guesses on all questions.
  1. In a game, a player flips a coin 20 times. The player wins if at least 15 heads appear. Find the probability of winning.
  1. A biased coin has a 0.9 probability of landing on heads. If the coin is tossed 8 times, find the probability of getting exactly 6 heads.

Solutions to Practice Problems

  1. $P(X \geq 3)$
  1. $E(X)$
  1. $P(X \geq 15)$
  1. $P(X = 6)$ These problems will help you apply the concepts of Bernoulli trials and binomial distribution.

Additional Resources

  • Textbook: “Probability and Statistics” by Sheldon Ross
  • Online course: “Introduction to Probability and Statistics” on Khan Academy
  • Websites: Stat Trek, MathIsFun These resources provide further explanations, examples, and exercises on the topic of binomial distribution.