Probability often changes the moment you learn that some related event has already happened. Conditional probability $P(A\mid B)$ is the probability that event $A$ occurs given that event $B$ has already occurred. Learning that $B$ happened shrinks the world of possibilities from the whole sample space down to $B$ alone, so we measure $A$ only against that reduced space:
$$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)},\qquad P(B)>0.$$
The numerator counts outcomes that are in both $A$ and $B$; the denominator rescales by the new, smaller universe $B$. If $P(B)=0$ the conditional probability is undefined — you cannot condition on something impossible.
Properties of conditional probability. For a fixed conditioning event $B$ with $P(B)>0$, the function $A\mapsto P(A\mid B)$ behaves like an ordinary probability measure:
- $0\le P(A\mid B)\le 1$, and $P(S\mid B)=P(B\mid B)=1$ (the whole space, or $B$ itself, is certain once $B$ is given).
- If $A$ and $C$ are mutually exclusive, then $P(A\cup C\mid B)=P(A\mid B)+P(C\mid B)$.
- $P(A'\mid B)=1-P(A\mid B)$ — the complement rule still holds inside the conditioned world.
Multiplication theorem of probability
Rearranging the definition gives the probability that both events occur. Because $A\cap B$ can be read two ways, there are two equivalent forms:
$$P(A\cap B)=P(B)\,P(A\mid B)=P(A)\,P(B\mid A).$$
This is the engine behind "draw without replacement" problems: the probability of a sequence of dependent events is the product of one unconditional probability and the conditional probabilities that follow. For three events it extends naturally to $P(A\cap B\cap C)=P(A)\,P(B\mid A)\,P(C\mid A\cap B)$.
Independent events
Two events are independent when the occurrence of one does not change the probability of the other. The cleanest test is the product rule:
$$A,B \text{ independent}\iff P(A\cap B)=P(A)\,P(B).$$
Equivalently $P(A\mid B)=P(A)$ and $P(B\mid A)=P(B)$ — conditioning tells you nothing new. A useful consequence: if $A$ and $B$ are independent, then so are the pairs $A'$ and $B$, $A$ and $B'$, and $A'$ and $B'$.
Mutually exclusive is NOT the same as independent. This is the single most common trap in the chapter. Mutually exclusive (disjoint) events cannot happen together, so $P(A\cap B)=0$; independent events satisfy $P(A\cap B)=P(A)P(B)$. If both $P(A)>0$ and $P(B)>0$, these two conditions contradict each other — so two events with non-zero probability that are mutually exclusive are necessarily dependent (knowing one happened tells you the other did not).
Deeper Insight — conditioning is "updating", independence is "no update". Every formula in this topic is one idea seen from different angles: probability is a measure of belief, and a conditional probability is that belief after new information arrives. The multiplication theorem just runs the definition backwards to build joint probabilities step by step, which is exactly how you compute the odds of a chain of dependent draws. Independence is the special, convenient case where the new information is irrelevant, so the update does nothing and the joint probability collapses to a plain product. Hold on to the distinction between "disjoint" (the events fight each other) and "independent" (the events ignore each other) and most of the chapter's word problems sort themselves out.
A die is rolled. Find $P(\text{number}>3 \mid \text{number is even})$.
SolutionLet $A$ be "number $>3$" and $B$ be "number is even". Even outcomes are $\{2,4,6\}$, so $P(B)=\tfrac36$. Outcomes that are both even and $>3$ are $\{4,6\}$, so $P(A\cap B)=\tfrac26$. Therefore $P(A\mid B)=\dfrac{2/6}{3/6}=\dfrac{2}{3}.$ Answer: $\dfrac{2}{3}$.
If $P(A)=0.5,\ P(B)=0.4,\ P(A\cap B)=0.2$, are $A,B$ independent?
SolutionApply the product test. $P(A)P(B)=0.5\times0.4=0.2$, which equals the given $P(A\cap B)=0.2$. Since $P(A\cap B)=P(A)P(B)$, the events satisfy the independence condition. Answer: Yes, $A$ and $B$ are independent.
Two cards are drawn without replacement from a well-shuffled deck. Find $P(\text{both kings})$.
SolutionThis is a dependent draw, so use the multiplication theorem. The first card is a king with probability $\tfrac{4}{52}$. Given a king has been removed, $3$ kings remain among $51$ cards, so the second is a king with probability $\tfrac{3}{51}$. Hence $P=\dfrac{4}{52}\times\dfrac{3}{51}=\dfrac{12}{2652}=\dfrac{1}{221}.$ Answer: $\dfrac{1}{221}$.
If $P(A)=0.6,\ P(B\mid A)=0.5$, find $P(A\cap B)$.
SolutionBy the multiplication theorem $P(A\cap B)=P(A)\,P(B\mid A)=0.6\times0.5=0.3.$ Answer: $0.3$.
If $P(A)=\tfrac{7}{13},\ P(B)=\tfrac{9}{13}$ and $P(A\cap B)=\tfrac{4}{13}$, find $P(A\mid B)$.
SolutionUse the definition directly: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{4/13}{9/13}=\dfrac{4}{9}.$ The $13$s cancel because both probabilities share the same denominator. Answer: $\dfrac{4}{9}$.
A coin is tossed three times. Given that the first toss is a head, find the probability of getting exactly two heads in all.
SolutionCondition on the first toss being a head; the reduced sample space is the four equally likely outcomes $\{HHH, HHT, HTH, HTT\}$. "Exactly two heads in all" requires exactly one more head in the last two tosses: $HHT$ and $HTH$. So the probability is $\dfrac{2}{4}=\dfrac12.$ Answer: $\dfrac{1}{2}$.
A problem is given to three students whose chances of solving it are $\tfrac12,\ \tfrac13$ and $\tfrac14$, independently. Find the probability that the problem is solved.
SolutionIt is easiest to find the probability that none solves it and subtract from $1$. The chances of failing are $\tfrac12,\ \tfrac23,\ \tfrac34$. By independence, $P(\text{none solves})=\tfrac12\times\tfrac23\times\tfrac34=\tfrac{6}{24}=\tfrac14.$ Therefore $P(\text{solved})=1-\tfrac14=\tfrac34.$ Answer: $\dfrac{3}{4}$.
Events $A$ and $B$ are such that $P(A)=\tfrac12,\ P(B)=\tfrac{7}{12}$ and $P(\text{not }A\text{ or not }B)=\tfrac14$. State whether $A$ and $B$ are independent.
SolutionFirst, $P(A'\cup B')=\tfrac14$. By De Morgan, $A'\cup B'=(A\cap B)'$, so $P(A\cap B)=1-\tfrac14=\tfrac34$. But $P(A\cap B)$ cannot exceed $P(A)=\tfrac12$, so the data already forces a check — recompute against the product $P(A)P(B)=\tfrac12\times\tfrac{7}{12}=\tfrac{7}{24}$. Since $\tfrac34\ne\tfrac{7}{24}$, $P(A\cap B)\ne P(A)P(B)$. Answer: $A$ and $B$ are not independent.
Show that two events $A$ and $B$ with $P(A)=0.3$ and $P(B)=0.4$ cannot be both mutually exclusive and independent.
SolutionIf they were mutually exclusive, $P(A\cap B)=0$. If they were independent, $P(A\cap B)=P(A)P(B)=0.3\times0.4=0.12$. These two values disagree ($0\ne0.12$), so both conditions cannot hold at once. Because both probabilities are non-zero, the events here are in fact mutually exclusive or independent, never both. Answer: Impossible — mutually exclusive events with non-zero probability are dependent.
A bag contains $5$ red and $3$ black balls. Two balls are drawn one after another without replacement. Find the probability that the first is red and the second is black.
SolutionUse the multiplication theorem for dependent events. $P(\text{1st red})=\tfrac58$. After removing one red, $7$ balls remain of which $3$ are black, so $P(\text{2nd black}\mid \text{1st red})=\tfrac37$. Hence $P=\tfrac58\times\tfrac37=\dfrac{15}{56}.$ Answer: $\dfrac{15}{56}$.
If $A$ and $B$ are independent events with $P(A)=0.3$ and $P(B)=0.6$, find (a) $P(A\cap B)$, (b) $P(A\cup B)$, and (c) $P(A'\cap B')$.
Solution(a) Independence gives $P(A\cap B)=0.3\times0.6=0.18$. (b) $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.3+0.6-0.18=0.72$. (c) $A'$ and $B'$ are also independent, so $P(A'\cap B')=P(A')P(B')=0.7\times0.4=0.28$ (equivalently $1-P(A\cup B)$). Answer: (a) $0.18$, (b) $0.72$, (c) $0.28$.