Probability

What Probability Measures

The probability of an event is a number between $0$ and $1$ that measures how likely the event is. When every outcome in a sample space is equally likely, probability is a ratio of counts:

$P(E) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$

A probability of $0$ means the event is impossible; a probability of $1$ means it is certain. On the GRE, the traps rarely come from this definition — they come from combining events, from the phrase "at least one," and from conditional language. Master a handful of rules and you defuse almost every Data Analysis probability question.

The Complement Rule

The complement of an event $E$, written $E^{c}$, is "$E$ does not happen." Because $E$ either happens or it does not,

$P(E^{c}) = 1 - P(E)$

The complement is the single most powerful tool in GRE probability. Whenever a question asks for the probability of "at least one" of something, computing the complement — the probability of none — is almost always faster than adding up cases.

The Addition Rule (Union)

For the probability that event $A$ or event $B$ occurs, use the general addition rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

The subtraction removes the overlap you would otherwise double-count. Two events are mutually exclusive (disjoint) when they cannot both occur, so $P(A \cap B) = 0$ and the rule collapses to $P(A) + P(B)$. Rolling a $2$ and rolling a $5$ on one die are mutually exclusive; drawing a card that is a king and a card that is a heart are not (the king of hearts is both).

The Multiplication Rule (Intersection)

For the probability that both $A$ and $B$ occur, multiply:

$P(A \cap B) = P(A) \cdot P(B \mid A)$

If the events are independent — one occurring does not change the probability of the other — this simplifies to $P(A) \cdot P(B)$. Successive coin flips are independent; drawing two cards without replacement are not, because the first draw changes the deck.

Situation	Rule
$A$ or $B$, disjoint	$P(A) + P(B)$
$A$ or $B$, overlapping	$P(A) + P(B) - P(A \cap B)$
$A$ and $B$, independent	$P(A) \cdot P(B)$
$A$ and $B$, dependent	$P(A) \cdot P(B \mid A)$

"At Least One" — Almost Always the Complement

Suppose you flip a fair coin $4$ times and want the probability of at least one head. Listing the cases (exactly one, exactly two, …) is slow. The complement — no heads — is a single product:

$P(\text{no heads}) = \left(\tfrac{1}{2}\right)^4 = \tfrac{1}{16}, \qquad P(\text{at least one head}) = 1 - \tfrac{1}{16} = \tfrac{15}{16}$

Whenever you see "at least one," reach for $1 - P(\text{none})$ first.

Conditional Probability

The conditional probability of $A$ given that $B$ has occurred is

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

In words: restrict the sample space to the outcomes where $B$ happened, then ask what fraction of those also have $A$. Conditional questions often read "given that…," "among the…," or "if we know that…." The key discipline is to change your denominator to the conditioning group.

Geometric Probability

Sometimes outcomes are not countable but fill a region — a length, an area, or a time interval. Then probability is a ratio of measures:

$P = \frac{\text{favorable length (or area)}}{\text{total length (or area)}}$

If a point is chosen at random inside a $10 \times 10$ square and you want the chance it lands inside an inscribed circle of radius $5$, the answer is $\dfrac{\pi (5)^2}{10^2} = \dfrac{25\pi}{100} = \dfrac{\pi}{4}$.

A Reliable Game Plan

• "At least one"? Compute $1 - P(\text{none})$.
• "$A$ or $B$"? Add, then subtract the overlap (zero if disjoint).
• "$A$ and $B$"? Multiply; ask whether the events are independent before you use $P(A)\cdot P(B)$.
• "Given that…"? Shrink the denominator to the conditioning event.
• Random point in a region? Take a ratio of areas or lengths.