Probability and Conditional Probability

Probability as a Fraction of Outcomes

The probability of an event is the fraction of outcomes that count as favorable, assuming every outcome is equally likely:

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

Every probability lands between $0$ (impossible) and $1$ (certain), and can be written as a fraction, decimal, or percent. On the SAT the "total" is almost always a count you read from a table or a described group, so the entire task collapses to one decision: identify the right numerator and the right denominator. Get those two counts right and the arithmetic is trivial.

Compound Events: "And" vs. "Or"

Two events combine in predictable ways.

"And" (both happen). For independent events — where one outcome does not affect the other — multiply: $P(A \text{ and } B) = P(A) \cdot P(B)$ Rolling a $6$ and flipping heads combines by multiplication because the die and the coin ignore each other.

"Or" (at least one happens). For events that cannot occur together (mutually exclusive), add: $P(A \text{ or } B) = P(A) + P(B)$ But when events can overlap, you must subtract the part counted twice: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ Forgetting that subtraction is the classic "or" mistake.

Drawing Without Replacement

A subtle case is drawing items without replacement, where the denominator shrinks after each draw. If a drawer holds $5$ black and $3$ white socks, the chance of pulling two black socks in a row is $\frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ The second fraction uses $4$ black out of $7$ remaining, because the first sock is already gone. If the item is replaced, both fractions stay the same and the draws are independent.

Complements — the Shortcut for "At Least One"

Sometimes the fastest path is the complement: $P(\text{not } A) = 1 - P(A)$ "At least one" problems are the signal. Computing the probability of at least one success directly can mean adding several cases, but the complement — zero successes — is usually a single product. Find $P(\text{none})$ and subtract from $1$.

Conditional Probability and Two-Way Tables

Conditional probability asks for the chance of one event given that another has already occurred. The notation $P(A \mid B)$ reads "the probability of $A$ given $B$," and the defining move is that the condition restricts the denominator to only the group where $B$ is true:

$P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)} = \frac{\text{count in both } A \text{ and } B}{\text{count in } B}$

The SAT delivers these almost exclusively through two-way tables. Consider a survey of students about whether they play a musical instrument:

	Plays instrument	Does not	Total
Juniors	48	32	80
Seniors	27	63	90
Total	75	95	170

Question: If a student is chosen at random from those who play an instrument, what is the probability the student is a junior?

The phrase "from those who play an instrument" restricts the pool to the Plays instrument column, whose total is $75$. That $75$ is the denominator — not the grand total of $170$. Among instrument players, $48$ are juniors, so $P(\text{junior} \mid \text{plays}) = \frac{48}{75} = \frac{16}{25}$

Now compare a question with no condition — "what is the probability a random student is a junior and plays an instrument?" The denominator returns to the full $170$: $\tfrac{48}{170} = \tfrac{24}{85}$. The word "given" is the switch that changes the denominator, and confusing these two denominators is the single most common error on the entire topic.

Order Matters in Conditionals

$P(A \mid B)$ and $P(B \mid A)$ are usually different numbers because they divide by different totals. In the table above, $P(\text{senior} \mid \text{plays}) = \tfrac{27}{75}$, but $P(\text{plays} \mid \text{senior}) = \tfrac{27}{90}$. The same overlap count of $27$ sits over two different denominators. Reading the condition carefully — which group are we inside? — settles which total to use.

Common Mistakes to Avoid

• Using the grand total as the denominator when the question says "given" — the condition restricts the group.
• Adding probabilities for "and," or multiplying for "or."
• Forgetting the denominator shrinks when drawing without replacement.
• Ignoring overlap in "or" problems and double-counting outcomes that satisfy both events.