Week 4 — Probability Rules & Random Variables (Intro)

The student government’s fall fundraiser sets up a table outside the union: a $2 raffle ticket next to it, and the campus coffee cart’s “lucky card” promotion — draw one card from a shuffled deck, win a free drink if it’s a heart. Both games raise the same question a smart customer should ask before playing: what are my actual chances, and what should I expect to walk away with? Last week you learned to read probability as long-run relative frequency and met the complement rule. This week adds two more building-block rules — for combining probabilities with “or” and with “and” — and introduces a new tool, the random variable, that finally lets you answer the second question: not just “will I win?” but “what’s this worth, on average?”

1Combining events with “or”: the addition rule¶

Every probability lives on the probability scale: a number from 0 (the event never happens) to 1 (it always happens) — you’ll see this written as a decimal or a percent. Recall from last week that the complement of an event $A$, written $A^c$, is “$A$ does not happen,” and $P(A^c) = 1 - P(A)$. This week’s first new idea is combining two events with the word “or.” The union of events $A$ and $B$ — written $A \text{ or } B$ — is the event that at least one of them happens.

If $A$ and $B$ can never happen on the same outcome, they’re called disjoint (or mutually exclusive), and $P(A \text{ and } B) = 0$. For disjoint events, probabilities simply add:

If $A$ and $B$ can happen together, adding $P(A)+P(B)$ double-counts the outcomes where both occur, so the general addition rule subtracts that overlap once:

Here $P(A \text{ and } B)$ is the joint probability that both events happen at once.

Worked Example 1 — the coffee cart’s lucky-card game. The cart uses a standard, well-shuffled 52-card deck; a customer draws one card, and every card is equally likely ($1/52$ each). Let $A$ = “the card is a heart” and $B$ = “the card is a face card” (jack, queen, or king, in any suit). There are 13 hearts, so $P(A) = 13/52 = 0.25$. There are 12 face cards, so $P(B) = 12/52 \approx 0.2308$.

Complement. The probability of not drawing a heart: $P(A^c) = 1 - 0.25 = 0.75$ (75%).

Disjoint case. “Heart or spade” — a single card can’t be both suits at once, so these events are disjoint, and $P(A \text{ and spade}) = 0$:

General case. “Heart or face card” is different — the jack, queen, and king of hearts belong to both events, so $P(A \text{ and } B) = 3/52 \approx 0.0577$. Simply adding $P(A)+P(B)$ would count those three cards twice, so the general rule subtracts the overlap:

About 42.3% of draws win at least one prize.

Independence and the multiplication rule. Two events are independent if knowing the outcome of one tells you nothing about the other — like two different customers each drawing from their own separately shuffled deck. For independent events, the probability both happen is the product:

So the probability that two independent customers both draw hearts is $0.25 \times 0.25 = 0.0625$ (6.25%).

The same rule governs any two independent processes, even something as ordinary as two coin flips. Simulating 1000 pairs of independent fair-coin flips (set.seed(1209)) landed both heads 261 times out of 1000 — an empirical probability of $261/1000 = 0.261$ — close to the theoretical $0.5 \times 0.5 = 0.25$ the multiplication rule predicts; the small gap is exactly the random noise you’d expect from only 1000 repetitions.

Figure 1. A bar chart of 1000 simulated two-coin-flip repetitions shows three bars — 0 heads, 1 head, 2 heads — at heights 243, 496, and 261 out of 1000. The middle bar (exactly one head) is roughly twice as tall as each outer bar, matching the lopsided-in-the-middle shape the multiplication rule predicts for two independent coins.

2Random variables and probability distributions¶

A random variable, usually written $X$, assigns a number to the outcome of a random process — dollars won, number of heads, minutes late. A discrete random variable takes on a countable list of specific values (a continuous random variable, coming in two weeks, can take any value in a range). The probability distribution of $X$ is a table listing every possible value $x$ alongside $P(x)$, the probability $X$ equals that value. Because the table lists every possibility and no two rows overlap, a valid distribution always satisfies $\sum P(x) = 1$.

Worked Example 2 — the fundraiser raffle. The raffle sells $n=500$ tickets at $2 each. Prizes: 1 grand prize of $250, 4 prizes of $50, and 20 prizes of $10; the remaining tickets win nothing. Let $X$ = the dollar amount a randomly selected ticket wins:

The probabilities sum to exactly 1, confirming this is a valid distribution.

Expected value. The expected value of a discrete random variable is the probability-weighted average of its possible values:

Here $\sum$ means “add up every row,” $x$ is each possible value, and $P(x)$ is its probability. $E(X)$ (also written $\mu$, “mu”) is the long-run average — not what any single ticket wins, but what the average payout per ticket would settle toward if the same raffle ran over and over. Multiplying and adding row by row:

So $E(X) = \$1.30$. Since a ticket costs $2, the expected net result of playing is $1.30 - 2.00 = -\$0.70$ per ticket — on average, a player loses about 70 cents per ticket, which is exactly the money that funds the student-government fundraiser. No individual ticket ever actually wins $1.30; the number describes the average over many, many tickets.

Figure 2. A probability-distribution bar chart for the raffle has four bars at $x=0,10,50,250$; the bar at $x=0$ towers to a height of 0.95, while the other three bars are barely visible slivers at 0.04, 0.008, and 0.002 — a strongly lopsided distribution where “win nothing” dominates, even though the rare big prize is what pulls $E(X)$ up above zero.

See it in R.

suppressMessages({library(mosaic); library(BSDA)})

# Worked Example 1: probability rules on a standard 52-card deck
p_heart <- 13/52
p_face  <- 12/52
p_heart_and_face <- 3/52                                   # J, Q, K of hearts
p_heart_or_face  <- p_heart + p_face - p_heart_and_face    # general addition rule
p_heart_or_spade <- 13/52 + 13/52                          # disjoint -> simple addition
p_not_heart <- 1 - p_heart                                 # complement rule
p_heart^2                                                   # multiplication rule, independent

# empirical check of the multiplication rule: 1000 pairs of independent fair coins
set.seed(1209)
sim <- do(1000) * rflip(2)
tally(~ heads, data = sim)
mean(~ (heads == 2), data = sim)     # empirical P(both heads)

# Worked Example 2: raffle distribution table and expected value
x <- c(250, 50, 10, 0)
p <- c(1/500, 4/500, 20/500, 475/500)
sum(p)            # check the distribution sums to 1
sum(x * p)         # E(X)

Running tally(~ heads, data = sim) prints exactly the 243 / 496 / 261 counts behind Figure 1, and sum(x * p) returns 1.3, confirming $E(X) = \$1.30$ from Example 2.

3Check your understanding¶

1. Using last week’s shelter data, $P(\text{Large}) = 67/240 \approx 0.279$. What is $P(\text{not Large})$, the probability a randomly chosen intake is not a Large animal? Which rule did you use?

2. A fair six-sided die is rolled once. Are “rolling a 2” and “rolling a 5” disjoint events? Find $P(\text{2 or 5})$.

3. Using the coffee cart’s deck (Worked Example 1), find $P(\text{diamond or ace})$. Are these two events disjoint? Show the general addition rule step you used.

4. Recall from Week 3’s coffee_sim data that $P(\text{Latte}) \approx 0.313$ (94 of 300 orders). If two different customers place their orders independently, what is the probability both order a latte?

5. A small carnival ring-toss game charges to play. Let $X$ = dollars a player wins, with this distribution:

   $x$	$5	$1	−$1
   $P(x)$	0.10	0.30	0.60

   (a) Verify this is a valid probability distribution. (b) Find $E(X)$ and interpret it in one plain-language sentence.

6. Worked Example 2 found $E(X) = \$1.30$ for the raffle, even though no single ticket can actually win exactly $1.30. Explain what $E(X) = \$1.30$ means, using the phrase “long-run average.”

4Key terms¶

• Probability scale — probability is always a number from 0 (never happens) to 1 (always happens).

• Complement ($A^c$) — the event “$A$ does not happen”; $P(A^c) = 1 - P(A)$.

• Union (“or”) — the event that at least one of $A$ or $B$ happens.

• Disjoint (mutually exclusive) events — events that cannot both happen on the same outcome, so $P(A \text{ and } B) = 0$.

• Addition rule (disjoint) — $P(A \text{ or } B) = P(A) + P(B)$, valid only when $A$ and $B$ are disjoint.

• Addition rule (general) — $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$; the subtraction removes double-counted outcomes.

• Joint probability, $P(A \text{ and } B)$ — the probability that both $A$ and $B$ happen.

• Independent events — knowing whether one event happened gives no information about the other.

• Multiplication rule (independent events) — $P(A \text{ and } B) = P(A) \times P(B)$.

• Random variable ($X$) — a variable that assigns a number to the outcome of a random process.

• Discrete probability distribution — a table of every possible value $x$ of a discrete random variable with its probability $P(x)$; a valid table always has $\sum P(x) = 1$.

• Expected value ($E(X)$, or $\mu$) — $\sum x \cdot P(x)$; the probability-weighted average value of $X$, interpreted as its long-run average over many repetitions.