Probability Distribution

A probability distribution may be defined as a function or table of values for a particular random variable mapping the outcomes in the sample space to the probabilities of those outcomes. It sounds more complicated than it is.

For instance, let's say we want a probability distribution for how many times we get tails if we toss a coin twice. The sample space would be: $\left\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\right\}$ where H represents heads and T represents tails.

Each outcome has a $\frac{1}{2}$ probability. Since we're looking for tails, we'll let X represent the random variable of how often we get tails. We're tossing the coin twice, so we could get tails zero times, one time, or two times. Here is the probability distribution for this experiment in table form:

In this article, we'll apply probability distributions to more complicated experiments and learn more about how they work. Let's get going!

Two types of probability distribution

Probability distributions can be classified into one of two types. The first type, discrete probability distribution, lists the probabilities of random variables with countable values. The binomial probability distribution is an excellent example of a discrete probability distribution.

The second type of probability distribution is called a continuous probability distribution or a probability density function. It lists the probabilities of random variables with values within a range and is continuous. The normal probability distribution is an example of a continuous probability distribution, and others are discussed in more advanced classes.

Thinking in terms of probability distribution

Now that we know the basics of probability distribution, it's time to put that knowledge to use. Let's say you take a math test with five multiple-choice questions, each with four answer choices. Unfortunately, you forgot to study, so you're taking a completely random guess on every item. What is the probability distribution for the number of correct answers?

Each problem has four total answer choices in this scenario, and only one is correct. That means you have a $\frac{1}{4}$ chance of guessing correctly and $\frac{3}{4}$ odds of answering incorrectly. The problem is asking us for correct answers, so that's our random variable and X. Possible values for X include 0, 1, 2, 3, 4, and 5. Therefore, we have a binomial probability distribution where $n=5$, $p=\frac{1}{4}$ , and $q=\frac{3}{4}$ . We know how to work with those!

All five answers are incorrect if we want to find $P\left(X=0\right)$. We can math out the probability of that happening as follows:

Next, let's consider the odds of answering one question correctly, $P\left(X=1\right)$. We'll have to use combinations since we don't care which question was answered correctly, just that a total of one was. The math looks like this:

We need to repeat this process for every possible value of X: meaning 2 correct answers, 3 correct answers, 4 correct answers, and 5 correct answers. The combinations get more complex each time, as there are only five different ways to answer one question correctly but 10 ways to get $C_2\frac{5}{2}=10$. Once we do all of that math, we'll end up with the following table:

Theoretically, you could score well on this test. Realistically, you should've studied. If your next math test is on probability distribution, hopefully you won't have to blindly guess on every question after solving this problem.

Practice problems

a. If a biased coin has a $\frac{2}{3}$ chance of landing on heads and $\frac{1}{3}$ of landing on tails, what is the chance that you need exactly five flips to get your first heads?

This problem is a binomial probability distribution where $p=\frac{2}{3}$ and $q=\frac{1}{3}$ . We care about the order here since we need four tails followed by heads, meaning we can solve this with a simple multiplication problem:

We can express our answer this way or simplify the expression and get an answer of about 0.008.

b. A fair coin is tossed 15 times. What is the probability of observing less than 3 heads?

This is a binomial probability distribution where $n=15$ and both p and q are $\frac{1}{2}$ (since the problem tells us we're working with a fair coin). $P\left(X=3\right)$ is the same as saying:

So, we're calculating all three of those values and then adding them together for our answer. Let's go in order:

Finally, we add the three sums to get our answer:

c. You roll a fair six-sided die five times. Create a probability distribution table for the number of sixes rolled.

The random variable we care about is how many sixes we roll, so that's X. We could roll six 0, 1, 2, 3, 4, or 5 times, so we need to use a binomial probability distribution to calculate the odds of each of those possibilities. Our p is $\frac{\mathrm{formula}}{6}$ since the die is fair, and q is 6. Let's start calculating.

Now that we have the odds for every possible scenario, we just have to list them all out in a table. Note that while some of these fractions can be reduced, we don't do so to make it easier to compare the odds we're looking at.

We're finished! Take your time on problems like this as even a slight error and dramatically throw off your results.

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