Cumulative probability distribution

Cumulative probability distributions

Understanding how probability accumulates across outcomes is central to working with random variables. The definitions below introduce two key concepts you’ll use throughout this section.

Definitions

Probability distribution

An equation, function, table, or graph that links the outcomes of a survey or another type of statistical experiment with the probability of each outcome. Cumulative probability distribution

Gives the probability that a random variable $X$ is less than or equal to a certain value.

Recall from Chapters 4.4 and 4.5 that software can calculate cumulative probabilities for geometric and binomial distributions efficiently:

• geometcdf $(p,x)$ gives the cumulative probability that the first success occurs on or before the $x$th trial, where $p$ is the probability of success on a single trial (geometric distribution).
• binomcdf $(n,p,x)$ gives the cumulative probability of $x$ or fewer successes in $n$ trials, where $p$ is the probability of success on a single trial (binomial distribution).

When you’re asked for a cumulative probability, the first step is to recognize the type of distribution involved.

• If the situation is binomial or geometric, you can use the appropriate software function.
• If the probabilities don’t follow a binomial or geometric pattern, you can still find the cumulative probability by adding the relevant individual probabilities.

For a discrete random variable, the general cumulative distribution function is:

$$F(x)=P(X\le x)=\sum _{k=0}^{x}P(X=k)$$

This general formula also works for geometric and binomial distributions. In those special cases, though, it’s usually faster to use the software functions instead of adding many individual probabilities.

Practice problems

Example 1:

Jared has a bowl of five number tiles. He randomly draws one tile with replacement (meaning he puts the tile back before the next draw). Two tiles are labelled $1$ point, one is labelled $2$ points, one is labelled $3$ points, and one is labelled $5$ points.

Part a.

What is the probability that Jared obtains exactly one point by drawing from the bowl once?

There are two tiles labelled $1$ point out of $5$ tiles total. So the probability of drawing exactly one point in a single draw is $\frac{2}{5}$.

Solution:

Part b.

What is the probability that Jared obtains less than three points?

“Less than three points” means the outcome is either $1$ point or $2$ points. These outcomes are mutually exclusive, so we add their probabilities:

$$\begin{array}{rl}P(X=1)& =\frac{2}{5}\\ P(X=2)& =\frac{1}{5}\\ P(X\le 2)& =P(X=1)+P(X=2)\\ & =\frac{2}{5}+\frac{1}{5}\\ & =\frac{3}{5}\end{array}$$

Solution:

Example 2:

Suppose Jared has a new bag of $10$ number tiles: two tiles labelled $1$ point, two tiles labelled $2$ points, three tiles labelled $3$ points, and three tiles labelled $4$ points.

Part a.

If Jared draws from the bag twice, what is the probability that he draws the same number of points both times?

There are four ways to draw the same number twice: two $1$s, two $2$s, two $3$s, or two $4$s. We find the probability of each and add them:

$$\begin{array}{rl}P(\text{both one point})& =\frac{2}{10}\cdot \frac{1}{9}=\frac{2}{90}\\ P(\text{both two points})& =\frac{2}{10}\cdot \frac{1}{9}=\frac{2}{90}\\ P(\text{both three points})& =\frac{3}{10}\cdot \frac{2}{9}=\frac{6}{90}\\ P(\text{both four points})& =\frac{3}{10}\cdot \frac{2}{9}=\frac{6}{90}\end{array}$$

Adding these together:

$$\begin{array}{rl}P(\text{same number})& =\frac{2}{90}+\frac{2}{90}+\frac{6}{90}+\frac{6}{90}\\ & =\frac{16}{90}\\ & =\frac{8}{45}\end{array}$$

Solution:

Part b.

If Jared draws from the bag twice, what is the probability that he draws different numbers each time?

You could list all the different-number outcomes, but it’s simpler to use the complement rule. The complement of “different numbers” is “the same number,” and Part a already found that probability:

$$\begin{array}{rl}P(\text{different numbers})& =1-P(\text{same number})\\ & =1-\frac{8}{45}\\ & =\frac{37}{45}\end{array}$$

Solution: