Conditional Probability

Information Changes Probability

Toss three fair coins: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$, each outcome with probability $\dfrac{1}{8}$. Let $E$ = "at least two heads" and $F$ = "first coin shows tail". Then $P(E) = \dfrac{4}{8} = \dfrac{1}{2}$.

Now suppose you are told the first coin showed tail. The outcomes with a first-coin head are ruled out: the world shrinks to $F = \{THH, THT, TTH, TTT\}$. Within this reduced world, only $THH$ has at least two heads, so the updated probability is $\dfrac{1}{4}$ — the information halved it.

This updated probability is the conditional probability of $E$ given $F$, written $P(E|F)$.

The definition

The favourable outcomes inside the reduced world $F$ are exactly the outcomes of $E \cap F$, so with equally likely outcomes $P(E|F) = \dfrac{n(E \cap F)}{n(F)}$, and dividing numerator and denominator by $n(S)$:

$$P(E|F) = \frac{P(E \cap F)}{P(F)}, \qquad P(F) \neq 0$$

Key Point: conditioning does not change the experiment — it changes the sample space. Every conditional question can be answered two ways: recount inside the reduced world $F$ (fast when outcomes are listable), or apply the formula (necessary when only probabilities are given). Both must agree.

The formula also works when outcomes are not equally likely — compute $P(E \cap F)$ and $P(F)$ by adding the individual outcome probabilities (tree diagrams organise this bookkeeping).

Properties of Conditional Probability

For events of a sample space $S$ with $P(F) \neq 0$:

1. Certainty given anything: $P(S|F) = P(F|F) = 1$ — the whole space, and $F$ itself, are certain once $F$ has occurred.

2. Addition rule survives conditioning:

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

and for disjoint $A, B$ the last term vanishes.

3. Complement rule survives too: $P(E'|F) = 1 - P(E|F)$.

Key Point: conditional probability $P(\cdot|F)$ behaves like an ordinary probability in its first slot — all the familiar rules hold with "$|F$" carried along. What it does NOT do is behave symmetrically: $P(E|F)$ and $P(F|E)$ are different numbers in general (they share the numerator $P(E \cap F)$ but divide by different things).

The classic traps

1. "Both boys given at least one boy" — the reduced world has three outcomes $\{bb, bg, gb\}$, not two: the answer is $\dfrac{1}{3}$, not $\dfrac{1}{2}$. Listing the reduced sample space prevents the error.
2. $P(A|B)$ vs $P(B|A)$: "probability the card is even given it exceeds 3" is not "probability it exceeds 3 given it is even". Read which event is given — it goes in the denominator.
3. Zero denominators: $P(A|B)$ is not defined when $P(B) = 0$ — there is no world to shrink to.

Solved Examples

Example 1: Straight from the formula

If $P(A) = \dfrac{7}{13}$, $P(B) = \dfrac{9}{13}$ and $P(A \cap B) = \dfrac{4}{13}$, evaluate $P(A|B)$.

Solution:

1. Formula: $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{4/13}{9/13}$.

Answer: $P(A|B) = \dfrac{4}{9}$ — note $P(A)$ itself was never needed.

Example 2: The two-children classic

A family has two children. What is the probability that both are boys, given that at least one is a boy?

Solution:

1. Sample space: $S = \{(b,b), (b,g), (g,b), (g,g)\}$.
2. Events: $E = \{(b,b)\}$; $F$ ("at least one boy") $= \{(b,b), (b,g), (g,b)\}$; $E \cap F = \{(b,b)\}$.
3. Formula: $P(E|F) = \dfrac{1/4}{3/4} = \dfrac{1}{3}$.

Answer: $\dfrac{1}{3}$ — the reduced world has three equally likely families, only one of them all-boy.

Example 3: A numbered-cards conditional

A card is drawn from ten cards numbered $1$ to $10$. Given that the number drawn is more than $3$, what is the probability that it is even?

Solution:

1. Events: $A$ = even $= \{2,4,6,8,10\}$; $B$ = greater than $3$ $= \{4,5,6,7,8,9,10\}$; $A \cap B = \{4,6,8,10\}$.
2. Count in the reduced world: $P(A|B) = \dfrac{n(A \cap B)}{n(B)} = \dfrac{4}{7}$.

Answer: $\dfrac{4}{7}$.

Example 4: Percentages as probabilities

A school has $1000$ students, $430$ of them girls; $10\%$ of the girls study in Class XII. What is the probability that a randomly chosen student studies in Class XII, given the student is a girl?

Solution:

1. Translate: $P(F) = 0.43$ (girl); $P(E \cap F) = 0.043$ (girl AND in Class XII, i.e. $10\%$ of $43\%$).
2. Formula: $P(E|F) = \dfrac{0.043}{0.43} = 0.1$.

Answer: $0.1$ — as it must be: "$10\%$ of the girls" is the conditional probability, and the formula simply recovers it.

Example 5: Dice with a known sum

A die is thrown twice and the sum is observed to be $6$. What is the conditional probability that $4$ has appeared at least once?

Solution:

1. Reduced world: $F = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}$ — five outcomes.
2. Favourable inside it: $E \cap F = \{(2,4), (4,2)\}$.
3. Count: $P(E|F) = \dfrac{2}{5}$.

Answer: $\dfrac{2}{5}$.

Example 6: Unequal outcome probabilities (a tree)

Toss a coin; if it shows head, toss again, and if tail, throw a die. Find the probability that the die shows a number greater than $4$, given there is at least one tail.

Solution:

1. Outcome probabilities (via the tree): $(H,H)$ and $(H,T)$ carry $\dfrac{1}{4}$ each; each $(T, i)$ carries $\dfrac{1}{12}$.
2. The events: $F$ = at least one tail $= \{(H,T), (T,1), \dots, (T,6)\}$ with $P(F) = \dfrac{1}{4} + 6 \times \dfrac{1}{12} = \dfrac{3}{4}$; $E \cap F = \{(T,5), (T,6)\}$ with probability $\dfrac{2}{12} = \dfrac{1}{6}$.
3. Formula: $P(E|F) = \dfrac{1/6}{3/4} = \dfrac{2}{9}$.

Answer: $\dfrac{2}{9}$ — with unequal outcomes, counting fails and only the probability form of the formula works.

Example 7: A question-bank conditional

A question bank has $300$ easy true/false, $200$ difficult true/false, $500$ easy multiple choice and $400$ difficult multiple choice questions. A question is selected at random. What is the probability that it is easy, given that it is multiple choice?

Solution:

1. Reduced world: multiple choice questions: $500 + 400 = 900$.
2. Favourable inside it: easy MCQs: $500$.
3. Count: $P(\text{easy}|\text{MCQ}) = \dfrac{500}{900} = \dfrac{5}{9}$.

Answer: $\dfrac{5}{9}$ — a two-way table question; conditioning just picks the MCQ column.