Probability Distribution of a Discrete Random Variable

From Chance Outcomes to a Probability Table

In What Is a Random Variable, you learned that a random variable assigns a number to each possible outcome of a chance process. In Discrete Versus Continuous Random Variables, you learned that a count such as the number of heads is discrete. Now we can describe a discrete random variable’s probability distribution: a list of its possible values together with the probability of each value.

Consider flipping a coin twice and counting the number of heads. The possible flip outcomes are \(HH\), \(HT\), \(TH\), and \(TT\), where the order records the first and second flips. The count is not the same thing as the ordered outcome: both \(HT\) and \(TH\) give one head. To find the probability of each count, we group together all outcomes that produce that count.

Here, define \(X\) as the number of heads in two flips. The variable can take the values 0, 1, or 2. Its probability distribution pairs each of those values with a probability. The notation \(P(X=1)\), for example, means the probability that the random variable \(X\) equals 1—that is, that exactly one head appears.

A useful way to build the distribution is to work in two stages: first identify the outcomes and their probabilities, then translate each outcome into the value of \(X\). When multiple outcomes give the same value, add their probabilities. This connects the distribution table to the addition rule for disjoint outcomes covered earlier in the course.

Building the Distribution for Two Fair Flips

Assume the coin is fair and the flips are independent. As in the earlier tutorials on independence and the multiplication rule for independent events, the probability of a complete pair of independent outcomes is the product of their individual probabilities. Each ordered outcome therefore has probability \(1/2 \times 1/2=1/4\).

The four ordered outcomes and their corresponding values of \(X\) are shown below. The outcome column keeps track of what happened on the flips; the \(X\) column records the number of heads.

There is one outcome that gives 0 heads, two outcomes that give 1 head, and one outcome that gives 2 heads. Because the outcomes are equally likely, the probabilities of the values are \(1/4\), \(2/4\), and \(1/4\), respectively.

The table can also be written with decimal probabilities: \(P(X=0)=0.25\), \(P(X=1)=0.50\), and \(P(X=2)=0.25\). Each entry refers to a value of \(X\), not to one particular ordered flip outcome. In particular, \(P(X=1)\) combines the probabilities of \(HT\) and \(TH\).

Reading Probabilities from the Distribution

Once the table is built, it becomes a compact way to answer questions about \(X\). For an event involving more than one value, add the probabilities of the values that meet the event’s description. For example, “at least one head” includes \(X=1\) and \(X=2\), while “no heads” includes only \(X=0\).

Be precise about the event. “Exactly one head” asks for the probability at one value, \(P(X=1)\). “At most one head” includes values 0 and 1, so it asks for \(P(X=0)+P(X=1)\). “More than one head” includes only \(X=2\) in this two-flip situation.

When the Coin Is Not Fair

The same method works when the possible outcomes are not equally likely. Suppose a coin has probability \(0.6\) of heads and \(0.4\) of tails on each flip, and the flips are independent. The ordered outcomes still are \(HH\), \(HT\), \(TH\), and \(TT\), but their probabilities are no longer all \(1/4\).

Multiply the probabilities along each outcome: \(P(HH)=0.6(0.6)=0.36\), \(P(HT)=0.6(0.4)=0.24\), \(P(TH)=0.4(0.6)=0.24\), and \(P(TT)=0.4(0.4)=0.16\). Since \(HT\) and \(TH\) both produce \(X=1\), add their probabilities to find \(P(X=1)\).

A Distribution Is a Model, Not a Record of One Trial

A probability distribution describes the long-run chance model for a random variable; it does not predict exactly what will happen on the next set of flips. In the fair-coin example, the model gives \(P(X=1)=0.50\), but a particular pair of flips has either one head or it does not. The table describes probabilities before observing the result.

This is different from a table of observed relative frequencies. If a student flips a coin twice once and gets one head, that single result does not establish that the probability of one head is 1. If many pairs of flips are recorded, the fraction with one head is an observed relative frequency. It may be compared with the model probability, but it is not the definition of the probability distribution.

The outcome-to-value mapping is an important part of the model. A different variable based on the same two flips could have a different distribution. For instance, a variable that records whether the first flip is heads takes only values 0 and 1; it is not the same variable as the number of heads across both flips. Always state what the random variable counts before constructing its table.

Common Mistakes and AP Exam Tips

• Listing outcomes as values of \(X\). \(HH\), \(HT\), \(TH\), and \(TT\) are outcomes, not values of the number-of-heads variable. The possible values of \(X\) are 0, 1, and 2.
• Forgetting that different outcomes can give the same value. Both \(HT\) and \(TH\) give one head. Add their probabilities to find \(P(X=1)\); do not list only one of them.
• Assuming all values have the same probability. The values 0, 1, and 2 are not equally likely for two fair flips. Count the outcomes that lead to each value, or add their probabilities.
• Adding probabilities for the wrong event. “Exactly one” means \(X=1\). “At least one” means \(X=1\) or \(X=2\). Translate the phrase into values before calculating.
• Using a product without the model’s assumptions. Multiplying flip probabilities is appropriate here because the flips are specified to be independent. If a chance process changes between trials, do not automatically use the same product.
• Writing a table without defining the variable. A full-credit response says what \(X\) represents, lists its possible values, and identifies the probability attached to each value.

For an AP response, make the mapping and probability reasoning visible. A complete explanation could say: “Let \(X\) be the number of heads in two flips. The outcomes \(HT\) and \(TH\) each have probability \(1/4\) and both give \(X=1\), so \(P(X=1)=1/4+1/4=1/2\).” This makes clear both why the outcomes are combined and how the table entry is calculated.

Key Takeaway

A probability distribution for a discrete random variable pairs every possible value with the probability of that value. To build one, define the variable, map each chance outcome to its value, and add probabilities when multiple outcomes produce the same value. Then use the table by adding the entries for values that meet the event’s description.