Probability Formula Solver

Every probability question reduces to picking the right rule. For a sample space $S$ of equally likely outcomes:

$P(A) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$

Complement rule — the fastest route to any "at least one" question:

$P(A^c) = 1 - P(A)$

Addition rule ("or"), which subtracts the double-counted overlap:

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

Multiplication rule ("and"), the general form:

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

Conditional probability, valid whenever $P(B) > 0$:

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

Every probability satisfies $0 \le P(A) \le 1$, and the probabilities of all outcomes in $S$ sum to exactly $1$.

The rules simplify only when a specific condition holds — checking that condition is the whole skill.

Condition	What it means	Simplified rule
Mutually exclusive	$A$ and $B$ cannot both occur, $P(A \cap B) = 0$	$P(A \cup B) = P(A) + P(B)$
Independent	$B$ happening does not change $P(A)$, so $P(A \mid B) = P(A)$	$P(A \cap B) = P(A)P(B)$

Mutually exclusive and independent are not the same thing — in fact two events with non-zero probability that are mutually exclusive are necessarily dependent, because one occurring drives the other's probability to zero.

Compound probability, in order

1. Translate the words: "or" $\to$ addition, "and" $\to$ multiplication, "at least one" $\to$ complement, "given that" $\to$ conditional.
2. Decide whether the trials are with or without replacement. With replacement keeps the events independent; without replacement changes the denominator on every draw.
3. Apply the general rule, then simplify only if the condition genuinely holds.
4. Sanity-check that the answer lies in $[0, 1]$.

The counting formula $P(A) = \text{favourable}/\text{total}$ assumes outcomes are equally likely. A loaded die or a biased coin breaks it, and you must then sum the individual outcome probabilities instead.

Frequent errors

• Adding without subtracting the overlap. If $A$ and $B$ can co-occur, $P(A) + P(B)$ counts the intersection twice.
• Multiplying dependent events as if independent. Drawing without replacement shrinks both the numerator and the denominator on the second draw.
• Reversing the condition. $P(A \mid B) \ne P(B \mid A)$ unless $P(A) = P(B)$; conflating them is the base-rate fallacy behind most medical-test puzzles.
• Computing "at least one" term by term. Use $1 - P(\text{none})$; the direct sum has many more cases and far more room to slip.
• Answering above 1. Any result outside $[0, 1]$ means a rule was misapplied, usually addition where multiplication was needed.