Topic 4.7 Introduction to Random Variables and Probability Distributions: define discrete random variables, represent and interpret their probability distributions, and use them to find probabilities of events.

What this topic is asking

The College Board (Topic 4.7) wants you to define a discrete random variable, represent and interpret its probability distribution, and use the distribution to find probabilities of events, including cumulative ("at least," "at most," "between") probabilities.

Random variables and their distributions

The shift from Topics 4.3 to 4.6 is that we now attach numbers to outcomes and study the whole pattern of those numbers. Instead of asking "what is the probability of this event?", we describe all the values a numerical outcome can take and how likely each is, the same descriptive move as Unit 1 (shape, center, spread), but now for a theoretical distribution rather than a dataset.

What makes a distribution valid

These two conditions are exam staples. Given all but one probability, the missing one is whatever makes the total $1$. Given a "distribution" whose probabilities sum to more or less than $1$, or include a negative value, you declare it invalid. This validity check is the first thing to do whenever a distribution is presented.

Finding probabilities of events

Once you have a valid distribution, the probability of any event is just the sum of the probabilities of the values that make up the event. The skill is translating the words into the right set of values. "$X$ is at least $2$" means $X \in \{2, 3, 4, \dots\}$, so add $P(2) + P(3) + \cdots$. "$X$ is at most $1$" means $\{0, 1\}$. "$X$ is between $1$ and $3$ inclusive" means $\{1, 2, 3\}$. The boundary words matter: "more than $2$" excludes $2$ ($\{3, 4, \dots\}$) while "at least $2$" includes it. A cumulative probability such as $P(X \le k)$ adds all values up to $k$, and the complement is often handy, $P(X \ge 2) = 1 - P(X \le 1)$, when the "tail" you want has more values than its opposite. Reading the inequality carefully and then summing the matching entries is the entire computational content of the topic, but it underlies everything that follows, because the mean, standard deviation, and the binomial and geometric distributions are all built on a probability distribution like this one.

Interpreting in context

As everywhere in AP Statistics, a probability is only fully answered when interpreted in context. $P(X \ge 2) = 0.6$ should be read as "there is a $60\%$ chance the dealership sells at least two cars on a given day," not left as a bare $0.6$. This habit matters more than it looks, because free-response scoring repeatedly awards a separate point for the contextual interpretation, and because it forces you to keep track of what the random variable actually counts or measures. A probability distribution is a model of a real quantity, daily car sales, number of defective items, points scored, and the point of computing its probabilities is to say something meaningful about that quantity. Pairing every numerical answer with a sentence in the situation's own terms is the discipline that turns a correct calculation into a complete response, and it carries directly into the expected-value and distribution topics next.

Try this

Q1. A distribution has $P(1) = 0.3$, $P(2) = 0.45$, $P(3) = k$. Find $k$. [1 point]

• Cue. Sum to one: $k = 1 - (0.3 + 0.45) = 0.25$.

Q2. For that distribution, find $P(X > 1)$ and interpret it. [2 points]

• Cue. $P(X > 1) = P(2) + P(3) = 0.45 + 0.25 = 0.70$; there is a $70\%$ chance $X$ exceeds $1$.