Probability Distributions (1) - Binomial Distribution

　A probability distribution is a function that expresses the probability of each outcome occurring in a given situation.

Function

Domain -> Range

Outcome -> Probability

　That’s roughly the relationship it represents. I’ll explain probability distributions in more depth another time.

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　Let’s look at one of the most representative probability distributions: the binomial distribution.

　Say that rolling a die once and getting a 1 counts as a success. On the first roll, the probability of success is 1/6. On the second roll, the probability of success is again 1/6. The probability of success on the second roll is 1/6 regardless of whether the first roll was a 1 or not. Let X be the number of successes as we keep rolling the die.

　In this example, suppose we want to find the probability of getting exactly 5 successes out of 10 rolls.

　The situation above has the following characteristics:

1. The same trial (rolling the die) is repeated.
2. Each trial results in either success (rolling a 1) or failure.
3. The probability of success p (1/6) is constant across trials.
4. Each trial is independent — the outcome of a previous trial doesn’t change the probability of later trials.

　We can use the binomial distribution in situations like this.

　For reference, this kind of situation is called a Bernoulli process, and each trial is called a Bernoulli trial.

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　I said we can use the binomial distribution — let’s see how.

　The number of successes X out of n trials is called a binomial random variable.

　The binomial distribution is the function that produces a probability from this binomial random variable.

　It’s denoted as follows.

$b(x; n, p)$

　Here, b is the function’s name (from “binomial”), x is the outcome (number of successes), n is the number of trials, and p is the probability of success on each trial.

　Now let’s derive the binomial distribution.

$b(x; n, p) = \binom{n}{x} \cdot p^x \cdot q^{n-x}$

　q is 1-p, and for anything else you don’t know, look up permutations and combinations.

　Now let’s solve the dice example from the start.

　In the problem, the number of trials n is 10, the number of successes x is 5, and the probability of success p is 1/6.

$b(x; n, p) = b\left(5; 10, \frac{1}{6}\right)$

$\binom{n}{x} \cdot p^x \cdot q^{n-x} = \binom{10}{5} \cdot \left(\frac{1}{6}\right)^5 \cdot \left(\frac{5}{6}\right)^{10-5} = 252 \cdot \frac{1}{7776} \cdot \frac{3125}{7776} = \frac{21875}{1679616}$

　That’s the probability of getting exactly 5 successes out of 10 dice rolls.

　When the numbers get messy, use a calculator or Wolfram Alpha.

Reference

Walpole, R. E., & Myers, R. H. (2016). Probability and statistics for engineers and scientists. New York: Macmillan.

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　This post was originally published on Naver Blog on February 29, 2020.