How New Information Changes the Probability of an Event

The chance that a card drawn from a pack is a king is $\frac{1}{13}$ — but if you are told it is a face card, the chance jumps to $\frac{1}{3}$. Updating a probability once you learn that something has happened is conditional probability, and it leads straight to the idea of independent events.

This part covers conditional probability, the multiplication theorem, and independent events.

**The conditional probability of $E$ given that $F$ has occurred is $P(E|F)=\frac{P(E\cap F)}{P(F)}$ with $P(F)\ne 0$; for equally likely outcomes it is $\frac{n(E\cap F)}{n(F)}$, because the sample space shrinks to $F$.

Worked example 1 — two coins.** Two coins are tossed. $E$: both heads; $F$: at least one head. Then $F=\{HH,HT,TH\}$ and $E\cap F=\{HH\}$:

$$P(E|F)=\frac{1/4}{3/4}=\frac{1}{3}$$

Worked example 2 — a die. A die is thrown. $E$: a number greater than $3$; $F$: an even number. Here $E\cap F=\{4,6\}$ and $F=\{2,4,6\}$:

$$P(E|F)=\frac{2}{3}$$

Properties, for $P(F)\ne 0$:

- $P(S|F)=P(F|F)=1$
- $P((A\cup B)|F)=P(A|F)+P(B|F)-P((A\cap B)|F)$
- $P(E^{\prime }|F)=1-P(E|F)$

An everyday example. The chance a train runs late is higher once you know there is thick fog — the fog is the condition that reshapes the probability.

The substance. **$P(E|F)$ and $P(F|E)$ usually differ** — in the coin example, $P(F|E)=1$ while $P(E|F)=\frac{1}{3}$.

**The multiplication theorem states $P(E\cap F)=P(F)P(E|F)=P(E)P(F|E)$, and for three events $P(E\cap F\cap G)=P(E)P(F|E)P(G|E\cap F)$.

Worked example 1 — without replacement.** An urn holds $10$ black and $5$ white balls. Two balls are drawn one after the other without replacement. The probability that both are black:

$$P=\frac{10}{15}\times \frac{9}{14}=\frac{90}{210}=\frac{3}{7}$$

Worked example 2 — three cards. Three cards are drawn without replacement from a well-shuffled pack. The probability that the first two are kings and the third is an ace:

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

Reading the steps. Each factor is a conditional probability — after one king is drawn, only $3$ kings remain among $51$ cards.

An everyday example. Picking two ripe mangoes in a row from a basket without putting the first back changes the odds for the second pick, exactly as the multiplication theorem accounts for.

The substance. With replacement the draws do not affect each other, so each conditional probability becomes an ordinary one.

**Events $E$ and $F$ are independent if $P(E\cap F)=P(E)P(F)$, so one occurring does not change the probability of the other; mutually exclusive events cannot occur together, so $P(E\cap F)=0$.

An equivalent test.** When $P(F)\ne 0$, independence means $P(E|F)=P(E)$.

Worked example 1 — independent. A die is thrown. $E$: a multiple of $3$, with $P(E)=\frac{1}{3}$; $F$: an even number, with $P(F)=\frac{1}{2}$. Then $E\cap F=\{6\}$:

$$P(E\cap F)=\frac{1}{6}=\frac{1}{3}\times \frac{1}{2}$$

so $E$ and $F$ are independent.

Worked example 2 — dependent. $E$: a number greater than $3$, with $P(E)=\frac{1}{2}$; $F$: an even number. Now $P(E\cap F)=\frac{2}{6}=\frac{1}{3}$, but $P(E)P(F)=\frac{1}{4}$, so they are dependent.

Independent versus mutually exclusive:

- Mutually exclusive — cannot happen together: $P(E\cap F)=0$
- Independent — can happen together, but neither affects the other: $P(E\cap F)=P(E)P(F)$
- Mutually exclusive events with non-zero probabilities are always dependent

Three events are mutually independent only if every pair and all three together satisfy the product rule.

An everyday example. Rain in Chennai and heads on a coin tossed in Delhi are independent, while one cricket match ending in a win and ending in a loss are mutually exclusive.

The substance. Independence must be checked with numbers — it cannot be decided from the wording of the events alone.