The average household has 1.25 pets

A random variable is a rule that attaches a numerical value to each outcome of a random process. Flip three coins and let $X$ be the number of heads; survey a household and let $X$ be the number of pets; play a game and let $X$ be the net dollars won.

A discrete random variable takes a countable set of values, typically from counting, and each value carries its own probability. A continuous random variable takes any value in an interval, typically from measuring, and probability is described by area under a density curve rather than by a value-by-value list. This topic works with discrete variables; the normal model in Topic 2.11 is the continuous case.

Capital letters name the variable and lowercase letters name a particular value, so $P(X = 2)$ reads as the probability that the random variable takes the value 2. The notation is worth keeping straight because it is what separates the variable from its values in every formula that follows.

The probability distribution of a discrete random variable gives every possible value along with its probability. Two requirements, inherited directly from the rules of probability, decide whether a table qualifies:

1. Every probability satisfies $0 \le P(x) \le 1$.
2. The probabilities sum to exactly 1, since the listed values cover every possible outcome.

Number of pets per household, $X$:

1. $P(0) = 0.30$, $P(1) = 0.35$, $P(2) = 0.20$, $P(3) = 0.10$, $P(4) = 0.05$.

Those sum to 1.00, and none is negative or above 1, so it is a valid distribution. A probability histogram draws it: one bar per value, the bar's height being the probability. Checking the sum first is the cheapest way to catch a missing category or a typo, and it is the check most often skipped.

Because the values are mutually exclusive, the probability of an event is the sum of the probabilities of the values in it:

1. $P(X \ge 2) = 0.20 + 0.10 + 0.05 = 0.35$.
2. $P(X \le 1) = 0.30 + 0.35 = 0.65$, which is the complement of the previous line.
3. $P(1 \le X \le 3) = 0.35 + 0.20 + 0.10 = 0.65$.
4. $P(X > 1) = 0.35$, since "more than 1" excludes 1 while "at least 1" includes it.

The boundary is where these go wrong. "At least 2" includes 2; "more than 2" does not; "at most 2" includes 2. For a discrete variable those are genuinely different sums, and the difference is exactly one bar's probability. Writing the list of values in the event before adding anything is what keeps it straight.

The expected value (or mean) of a discrete random variable is the probability-weighted average of its values:

$$\mu_X = E(X) = \sum x \cdot P(x).$$

For the pets distribution, $E(X) = 0(0.30) + 1(0.35) + 2(0.20) + 3(0.10) + 4(0.05) = 1.25$ pets. On the probability histogram it is the balance point, the same interpretation the mean carries for data.

Three readings of that number are wrong and common:

1. It is not a possible outcome. No household holds 1.25 pets, and nothing requires the expected value to be a value the variable can take. Rounding it to 1 discards information that later calculations need.
2. It is not the most likely value. That is the mode, here 1 pet at probability 0.35. Expected value and most likely coincide only by accident.
3. It is not a prediction about the next trial. The next household will have 0, 1, 2, 3, or 4 pets. The 1.25 describes the average across very many households.

The long-run reading is what makes expected value useful in decisions. A raffle ticket costing 5 dollars, with one 500 dollar prize among 200 tickets, has $E(\text{net}) = 500 \left(\frac{1}{200}\right) - 5 = -2.50$ dollars: over many tickets the buyer loses about 2.50 dollars per ticket on average, even though any single ticket returns either $-5$ or $+495$ dollars and never $-2.50$.