Why Knowing One Outcome Changes the Odds of Another

If a friend rolls two dice and tells you only that the first die shows an even number, the chance that the total is 8 is no longer what it was. Conditional probability measures exactly this — how the odds of one event shift once another event is known to have happened.

This lesson covers conditional probability, the multiplication theorem, independent and dependent events, and mixed problems.

**The conditional probability of A given B is $P(A|B)=\frac{P(A\cap B)}{P(B)}$, defined when $P(B)>0$ — it is the fraction of B's outcomes in which A also happens.

Why the formula works. Once B is known to have happened, B becomes the new sample space, so only outcomes inside B count.

Worked example.** Two dice are rolled. Let A be 'the sum is 8' and B be 'the first die is even'.

- B has $3\times 6=18$ outcomes
- Outcomes in both: $(2,6)$, $(4,4)$ and $(6,2)$, so 3 outcomes
- $P(A|B)=\frac{3/36}{18/36}=\frac{3}{18}=\frac{1}{6}$
- Without the information, $P(A)=\frac{5}{36}$, which is smaller

Worked example 2. If $P(A\cap B)=0.3$ and $P(B)=0.5$, then $P(A|B)=\frac{0.3}{0.5}=0.6$.

Properties. $P(S|B)=1$ and $P(A^{\prime }|B)=1-P(A|B)$.

An everyday example. The chance of rain in the evening given that dark clouds have gathered by afternoon is a conditional probability, and it is higher than the chance of rain on an ordinary day.

The substance. **$P(A|B)$ and $P(B|A)$ are usually different** — here $P(B|A)=\frac{3}{5}$, because 3 of the 5 ways to make a total of 8 start with an even number.

**The multiplication theorem states that $P(A\cap B)=P(A)P(B|A)=P(B)P(A|B)$ — the chance of both events is the chance of the first times the chance of the second once the first has happened.

Extension to three events.** $P(A\cap B\cap C)=P(A)P(B|A)P(C|A\cap B)$.

Worked example (without replacement). A bag has 5 red and 3 blue balls. Two are drawn one after the other without replacement. The probability that both are red is

$$P=\frac{5}{8}\times \frac{4}{7}=\frac{20}{56}=\frac{5}{14}\approx 0.357$$

Worked example (three draws). Three cards are drawn from a pack of 52 without replacement. The probability that all three are kings is

$$\frac{4}{52}\times \frac{3}{51}\times \frac{2}{50}=\frac{1}{5525}$$

With replacement. If the first ball is put back before the second draw, the draws no longer affect each other: $\frac{5}{8}\times \frac{5}{8}=\frac{25}{64}$.

An everyday example. Picking two class monitors one after another from a class of 40 changes the pool for the second pick, so the multiplication theorem with a conditional probability is needed.

The substance. The second factor must be conditional unless the events are independent — multiplying $\frac{5}{8}\times \frac{5}{8}$ for draws without replacement is a classic error.

**Events A and B are independent if and only if $P(A\cap B)=P(A)P(B)$, which is the same as saying $P(A|B)=P(A)$ — knowing that B happened does not change the chance of A.

Worked example (independent).** One die is rolled. Let A be 'even' $=\{2,4,6\}$ and B be 'a multiple of 3' $=\{3,6\}$.

- $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$
- $A\cap B=\{6\}$, so $P(A\cap B)=\frac{1}{6}$
- $\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}$, so A and B are independent

Worked example (dependent). Let C be 'greater than 3' $=\{4,5,6\}$. Then $A\cap C=\{4,6\}$ and $P(A\cap C)=\frac{1}{3}$, but $P(A)P(C)=\frac{1}{4}$, so A and C are dependent.

Useful results. If A and B are independent, so are A and $B^{\prime }$, $A^{\prime }$ and B, and $A^{\prime }$ and $B^{\prime }$.

An everyday example. Whether a coin lands heads and whether it rains in Pune the next day are independent — one tells you nothing about the other.

The substance. Independent is not the same as mutually exclusive — two mutually exclusive events with non-zero probabilities are always dependent, because if one happens the other cannot.

**Break the question into simple events, use the multiplication theorem for 'and', the addition theorem $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ for 'or', and the complement for 'at least one'.

Worked example (at least one).** Three students solve a problem independently with probabilities $\frac{1}{2}$, $\frac{1}{3}$ and $\frac{1}{4}$. The chance that none of them solves it is $\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}=\frac{1}{4}$, so the problem is solved with probability $1-\frac{1}{4}=\frac{3}{4}$.

Worked example (exactly one).

$$\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}+\frac{1}{2}\cdot \frac{1}{3}\cdot \frac{3}{4}+\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{1}{4}=\frac{6+3+2}{24}=\frac{11}{24}$$

Worked example (conditional with addition). If $P(A)=0.4$, $P(B)=0.3$ and $P(A\cap B)=0.12$, then $P(A\cup B)=0.58$, and the chance of A given that at least one of the events happened is $\frac{0.4}{0.58}\approx 0.69$.

An everyday example. A shop with two independent burglar alarms, each working with probability 0.9, is protected with probability $1-0.1\times 0.1=0.99$.

The substance. 'At least one' is almost always fastest through the complement — listing every case directly multiplies the chances of a slip.