Discrete Probability Distributions - Binomial Distribution

Discrete Probability Distributions

Binomial Random Variable

A random variable is binomial if it meets the following criteria:

  • It measures the number of successes in nn identical trials.
  • Each trial has only two possible outcomes: success or failure.
  • The probability of success (and failure) remains constant from one trial to another.
  • Trials are independent.

Example 1: Pat's Quiz

Pat, a statistics student, relies on guessing for a quiz with 10 multiple-choice questions, each having 5 possible answers with only 1 correct.

We want to find the probability that:

  • Pat answers all questions incorrectly.
  • Pat answers exactly 2 questions correctly.
  • Pat fails the quiz (score less than 50%, each question worth 1 point).

Example 2: Non-Binomial Scenario

Pat attempts to solve each question rather than guess. However, solving statistics questions is very brain-consuming and Pat gets tired, this means that the chance of getting each question right decreases as he works on the questions. This will not be a binomial experiment because the probability of success does not remain constant.

Bernoulli Distribution

Let BB be a random variable that takes the value 1 with probability π\pi and the value 0 with probability 1π1 - \pi.

  • Mean of BB:
    μB=E[B]=bP(b)=0×(1π)+1×π=π\mu_B = E[B] = \sum bP(b) = 0 \times (1 - \pi) + 1 \times \pi = \pi
  • Variance of BB:
    Var(B)=(bμB)2p(b)=(0π)2(1π)+(1π)2π=π2(1π)+(12π+π2)π=π2π3+π2π2+π3=ππ2=π(1π)\text{Var}(B) = \sum (b - \mu_B)^2 p(b) = (0 - \pi)^2 (1 - \pi) + (1 - \pi)^2 \pi = \pi^2 (1 - \pi) + (1 - 2\pi + \pi^2) \pi = \pi^2 - \pi^3 + \pi - 2\pi^2 + \pi^3 = \pi - \pi^2 = \pi(1 - \pi)

Binomial Distribution

Let XX be a random variable that follows the binomial distribution with parameters nn and π\pi, denoted as XBin(n,π)X \sim Bin(n, \pi).

  • XX is the sum of nn Bernoulli random variables with parameter π\pi:
    X=B<em>1+B</em>2++BnX = B<em>1 + B</em>2 + \cdots + B_n
  • Mean of XX:
    μ<em>X=E[X]=E[B</em>1]+E[B<em>2]++E[B</em>n]=nμB=nπ\mu<em>X = E[X] = E[B</em>1] + E[B<em>2] + \cdots + E[B</em>n] = n\mu_B = n\pi
  • Variance of XX:
    Var(X)=Var(B<em>1+B</em>2++B<em>n)\text{Var}(X) = \text{Var}(B<em>1 + B</em>2 + \cdots + B<em>n), where B</em>1,B<em>2,etc.B</em>1, B<em>2, \text{etc.} are independent of one another, so we can apply the rule about the variance of the sum of independent random variables: Var(X)=Var(B</em>1)+Var(B<em>2)++Var(B</em>n)=nπ(1π)\text{Var}(X) = \text{Var}(B</em>1) + \text{Var}(B<em>2) + \cdots + \text{Var}(B</em>n) = n\pi(1 - \pi)

Example: Number of Boys in a 6-Children Family

Let XX = number of boys in families with 6 children where n=6n = 6 and π=0.5\pi = 0.5 (assuming each birth has a 50% chance of being a boy).

Probability Distribution
xxp(x)p(x)
00.016
10.094
20.234
30.313
40.234
50.094
60.016
Sum1
Mean Calculation

μ=xp(x)=nπ\mu = \sum x p(x) = n \pi

In this example: μ=6×0.5=3\mu = 6 \times 0.5 = 3

xxp(x)p(x)xp(x)xp(x)
00.0160
10.0940.094
20.2340.468
30.3130.939
40.2340.936
50.0940.470
60.0160.096
Sum13.000
Variance Calculation

σ2=(xμ)2p(x)=nπ(1π)\sigma^2 = \sum (x - \mu)^2 p(x) = n \pi (1 - \pi)

In this example: σ2=6×0.5×(10.5)=1.5\sigma^2 = 6 \times 0.5 \times (1 - 0.5) = 1.5

xxp(x)p(x)xp(x)xp(x)xμ\bf{x - \mu}(xμ)2\bf{(x - \mu)^2}(xμ)2p(x)\bf{(x - \mu)^2 p(x)}
00.0160-390.144
10.0940.094-240.376
20.2340.468-110.234
30.3130.939000
40.2340.936110.234
50.0940.470240.376
60.0160.096390.144
Sum13.0001.500

Pat's Quiz - Population Mean and Standard Deviation

Suppose that a professor has a class of Pat. What is the population mean? What is the population standard deviation?

Binomial Distribution Table

Table 5 in Appendix B of the textbook provides binomial distribution probabilities.

Essentials of Statistics For Business and Economics, Anderson et al. 9th Edition, pages 813 to 821.