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3.3 Probability Distributions

Section Overview

Random Variables

What is a random variable? Let’s look at a 6-sided die. The die could land on any one of its six faces, but we don’t know for certain which face it will land on until after we’ve made our roll. This obviously relies on probability. We have:

P(X=k)P(X=k)

where:

XX = random variable

kk = specific value that the random variable takes on (in this case: 1, 2, 3, 4, 5, or 6)

The possible values that a random variables can take are determined by its probability distribution. In this section, we will learn about two common probability distributions: binomial distributions and geometric distributions.

Binomial Distribution

Binomial experiments, the experiments carried out by a binomial distribution, exhibit the following characteristics:

  1. The experiment consists of a fixed number, nn, of identical trials.

  2. Each trial results in one of two outcomes: success or failure

  3. The probability of success on a single trial is equal to some value pp and remains the same from trial to trial. The probability of a failure is equal to q=(1−p)q=(1-p).

  4. The trials are independent.

  5. A random variable YY is said to have a binomial distribution based on nn trials with success probability pp if and only if:

p(k)=(nk)pk(1−p)n−kp(k)=\binom{n}{k}p^k(1-p)^{n-k}

where

(nk)\binom{n}{k} = binomial coefficient

nn = number of trials

kk = number of successes

pp = probability of success

In general, if the random variable YY follows the binomial distribution with parameters nn and pp, we write Y∼Binomial⁡(n,p)Y\sim\operatorname{Binomial}(n,p). Remember not to mix up the order of the two parameters!

Example 3.3.1

A particular internship has a particular rejection rate of 20%. If 20 applicants apply:

a) Find the probability that exactly 14 be accepted.

b) Find the probability that more than 10 be accepted.

c) Find the probability that at least 16 be accepted.

Example 3.3.1 — recorded answer, introduction and part a

We are given the following information:

Let Y=the number of applicants who get the internshipY=\text{the number of applicants who get the internship}

We have 20 applicants total, so n=20n=20

The success probability is p=0.8p=0.8

The probability of success is the same for all 20 applicants, the number of applicants in the pool is a finite number, and there are only two possibilities either the applicants will be accepted or die, so we can use the binomial distribution function:

P(Y=y)=(nk)py(1−p)n−kP(Y=y)=\binom{n}{k}p^y(1-p)^{n-k}

a) We need to find out the probability that exactly 14 are accepted. That is, P(Y=14)P(Y=14)

But we know that YY has binomial distribution with parameter (20, 0.8)

P(Y=14)=(2014)(0.8)14(0.2)20−14P(Y=14)=\binom{20}{14}(0.8)^{14}(0.2)^{20-14}
P(Y=14)=38760(0.04398)(0.00064)P(Y=14)=38760(0.04398)(0.00064)
P(Y=14)=0.1091=10.91%P(Y=14)=0.1091=10.91\%
Source gap (S0242): The original answer says applicants are “accepted or die” and mixes y and k in its formula. The printed factor 0.00064 differs from the sixth power of 0.2, and the displayed intermediate product differs from the stated result 0.1091. The original wording, factors and result are retained.

The displayed calculator expressions:

binompdf⁡(20,0.80,14)=0.1091\operatorname{binompdf}(20,0.80,14)=0.1091
1−binomcdf⁡(20,0.80,12)=0.96791-\operatorname{binomcdf}(20,0.80,12)=0.9679
1−binomcdf⁡(20,0.80,15)=0.62961-\operatorname{binomcdf}(20,0.80,15)=0.6296
Example 3.3.1 — recorded answer, part c

c) If at least 16 applicants are accepted, then we calculate P(X≥16)P(X\ge16):

From what we know about the binomial cumulative density function from (b), we have:

P(X≥16)=1−P(X<16)P(X\ge16)=1-P(X<16)

We cannot easily calculate the above equation because the P(X<16)P(X<16) at the right of the equation doesn’t include the value of 16, which we need for the CDF. However, what is equivalent to “less than 16”? At most 15. Likewise, the equivalent value to “at least 16” is “more than 15”.

Rewriting our equation, and using our calculator (see Gallery 3.3.2 we get:

P(X>15)=1−P(X≤15)P(X>15)=1-P(X\le15)
P(X>15)=1−binomcdf⁡(20,0.80,15)=0.6296P(X>15)=1-\operatorname{binomcdf}(20,0.80,15)=0.6296

See Gallery 3.3.2 for how to calculate the probability on the calculator.

Example 3.3.1 — recorded answer, part b

b) If more than 12 applicants are accepted, then we calculate P(X>12)P(X>12). We are going to use the cumulative density function this time.

The binomial cumulative density function calculates all values up to and including a given random variable. Since P(X>12)P(X>12) includes P(X=13)+P(X=14)+⋯+P(X=20)P(X=13)+P(X=14)+\cdots+P(X=20), we want to subtract P(X≤12)P(X\le12) from 1, which includes

P(X=0)+P(X=1)+⋯+P(X=12):P(X=0)+P(X=1)+\cdots+P(X=12):
P(X>12)=1−P(X≤12)P(X>12)=1-P(X\le12)

From Gallery 3.3.2:

P(X>12)=1−binomcdf⁡(20,0.80,12)=0.9679P(X>12)=1-\operatorname{binomcdf}(20,0.80,12)=0.9679
Source gap (S0245): The source prints part c before part b on page 106. The prompt and gallery caption say “more than 10”, while the recorded part b work and calculator use “more than 12”; gallery frame 2 highlights bars beginning at 12. The original statements and frame are preserved without supplying a replacement answer.

Geometric Distribution

The random variable with the geometric probability distribution is associated with an experiment that shares many characteristics of a binomial experiment. However, now the random variable YY represents number of the trial on which the first success occurs. A random variable YY is said to have a geometric distribution if and only if:

p(k)=qk−1pp(k)=q^{k-1}p

where

kk = number of successes

pp = probability of success

qq = probability of failure

In general, if the random variable YY follows the goemetric distribution with parameters pp, we write Y∼Geometric⁡(p)Y\sim\operatorname{Geometric}(p).

Source gap (S0246): The original defines k as the “number of successes” after describing the trial on which the first success occurs. The definition is retained as recorded.
Example 3.3.2

Gold prospector will dig a succession of holes in a given area to find gold. The probability that he is successful on a given trial is 0.3.

a) What is the probability that the third hole dug is the first to yield gold?

b) What is the probability that it will take at most 10 holes to find a gold?

Example 3.3.2 — recorded answer

a) According to the problem, p=0.3p=0.3 and q=0.7q=0.7

If exactly 3 holes are dug, then we calculate:

p(3)=(0.7)3−1(0.3)=0.147p(3)=(0.7)^{3-1}(0.3)=0.147

Using the calculator (See Gallery 3.3.4), we also get:

P(X=3)=geometpdf⁡(0.30,3)=0.147P(X=3)=\operatorname{geometpdf}(0.30,3)=0.147

b) If at most 10 holes are dug, then we calculate P(X≤10)P(X\le10):

The geometric cumulative density function calculates all values up to and including a given random variable. Since P(X≤10)P(X\le10) includes P(X=10)+P(X=9)+⋯+P(X=0)P(X=10)+P(X=9)+\cdots+P(X=0)

Using the calculator (See Gallery 3.3.4), we get:

P(X≤10)=geometcdf⁡(0.30,10)=0.9717P(X\le10)=\operatorname{geometcdf}(0.30,10)=0.9717
Source gap (S0247): The answer references Gallery 3.3.4, while the calculator gallery is numbered 3.3.3. The text gives 0.9717, and the captured calculator screen gives 0.9718. Both values are retained.

The calculator screen displays:

geometpdf⁡(0.30,3)=0.147\operatorname{geometpdf}(0.30,3)=0.147
geometcdf⁡(0.30,10)=0.9718\operatorname{geometcdf}(0.30,10)=0.9718
Source gap (S0250): Gallery 3.3.4 is titled “Example 3.3.1” in the source, although it depicts the gold example 3.3.2. Its original title is retained.

Expectation and Variance

We’ve learned that summary measures of a distribution include its mean and its standard deviation/variance. Random variables also have a mean (sometimes called the expected value or expectation) and variance.

The mean is the average of the values of random variable XX each weighted by its probability:

μx=∑xipi=x1p1+x2p2+⋯+xnpn\mu_x=\sum x_i p_i=x_1p_1+x_2p_2+\cdots+x_np_n

The variance is calculated as thus:

σx2=∑(xi−μx)2pi=(x1−μx)2p1+⋯+(xn−μx)2pn\sigma_x^2=\sum(x_i-\mu_x)^2p_i=(x_1-\mu_x)^2p_1+\cdots+(x_n-\mu_x)^2p_n

Both binomial and geometric distributions have a distinct expectation (or mean) and variance:

DistributionMeanVariance
Binomialnpnpnp(1−p)np(1-p)
Geometric1/p1/pN/A

Pop Quiz 3.3.1

Question 1 of 5

You have a standard deck of 52 cards. You win $10 if the card drawn is an ace but lose $3 otherwise. What is the mean amount you’ll win?

Question 2 of 5

A binomial distribution has 8 trials with probability of success 0.25 on each trial. Use this data for the next two questions. What is the probability of exactly 2 successes?

Question 3 of 5

From question 2, what is the probability of at least 1 success?

Question 4 of 5

Suppose that there are 6 cans of vegetarian food and 3 cans of non-vegetarian food. 4 cans are selected when and a vegetarian can is selected on the second try. Let XX be the number of vegetarian cans chosen. XX is:

Question 5 of 5

An archer hits his target with probability 57\frac{5}{7}. He shoots at the target 12 times. Let XX be the number of misses in the 12 shoots. XX is:

Source gap (S0256): The original question 4 wording “4 cans are selected when and” and duplicate C/D choices in question 5 are retained. All five question prompts and choices are captured; no checked answer states are shown.
Glossary and deeper reading · 8 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Probability

    Probability quantifies how likely an event is under a specified chance model. It lies between 0 and 1, with larger values representing greater likelihood. For example, drawing a red card from a well-shuffled standard deck has probability one half.

    Read more: Probability in the captured Statistics reading →

  • Statistical independence

    Statistical independence means that learning one outcome does not change the probability distribution of another. For two events, it implies that the probability of both occurring equals the product of their probabilities. For example, independent draws with replacement keep the second draw’s color probabilities unchanged by the first.

    Read more: Statistical independence in the captured Statistics reading →

  • Binomial coefficient

    A binomial coefficient counts the ways to select a fixed number of objects from a set when order does not matter. In a binomial probability calculation, it counts the possible success positions among the trials. For example, choosing two success positions among four trials can be done in six ways.

  • Experiment

    An experiment deliberately imposes treatments and measures responses. Random assignment, comparison, and replication help distinguish treatment effects from other variation. For example, assigning plants to fertilizer treatments is an experiment.

    Read more: Experiment in the captured Statistics reading →

  • Mean

    The mean is a measure of center often called an average. For a finite data set, the arithmetic mean is the sum of the values divided by their number; for a random variable, the mean is its probability-weighted expected value when that value exists. Means can be pulled toward extreme values and are not generally resistant. For example, the arithmetic mean of 2, 4, and 9 is 5.

    Read more: Mean in the captured Statistics reading →

  • Standard deviation

    Standard deviation measures spread around the mean using squared deviations and a square root. A sample standard deviation uses the sum of squared deviations divided by one less than the sample size before taking the square root. For example, a larger standard deviation of travel times indicates greater variability, in the original time units.

    Read more: Standard deviation in the captured Statistics reading →

  • Variance

    Variance is the square of the standard deviation. It measures spread in squared units, with the precise averaging rule depending on whether it describes a sample or a probability distribution. For example, a standard deviation of 3 centimeters corresponds to a variance of 9 square centimeters.

  • Expected value

    Expected value is the probability-weighted mean of a random variable when that mean exists. It describes a long-run average rather than a value that must occur on one trial. For example, a fair die has expected value 3.5 even though no face is labeled 3.5.

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