Probability — chapter notes

If $E$ and $F$ are events of a sample space $S$ with $P(F) \ne 0$, the conditional probability of $E$ given $F$ is

Knowing that $F$ has happened rules out every outcome outside $F$; the outcomes inside $F$ stay equally likely. For equally likely outcomes, $P(E \mid F) = \dfrac{n(E \cap F)}{n(F)}$.

Example: two dice, given at least one six ($11$ outcomes), the probability that the sum is at least $10$ is $\dfrac{5}{11}$.

For a fixed event $F$ with $P(F) > 0$:

• $P(S \mid F) = P(F \mid F) = 1$
• $0 \le P(E \mid F) \le 1$
• $P(A \cup B \mid F) = P(A \mid F) + P(B \mid F) - P(A \cap B \mid F)$
• $P(E' \mid F) = 1 - P(E \mid F)$

In general $P(E \mid F) \ne P(F \mid E)$: the numerators are the same, the denominators are not. In a two-way table, divide by the row or column total of the given group.

From the definition,

For three events, $P(E \cap F \cap G) = P(E)\,P(F \mid E)\,P(G \mid E \cap F)$.

Use it for events in sequence, especially draws without replacement, where each later factor is conditional on the earlier draws. A tree diagram shows the stages: multiply along a path, add different paths.

Example: $10$ bulbs with $3$ fused, two drawn without replacement: $P(\text{both fused}) = \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{1}{15}$.

Events $E$ and $F$ are independent if $P(E \cap F) = P(E)\,P(F)$; equivalently (when the probabilities are non-zero) $P(E \mid F) = P(E)$ and $P(F \mid E) = P(F)$.

• If $E$, $F$ are independent, so are $E$ and $F'$, $E'$ and $F$, and $E'$ and $F'$.
• Three events are mutually independent if every pair satisfies the product rule and $P(E \cap F \cap G) = P(E)P(F)P(G)$.
• For independent events, $P(\text{at least one}) = 1 - P(E')P(F')\cdots$.
• Mutually exclusive ($P(E \cap F) = 0$) is a different idea: with non-zero probabilities, mutually exclusive events are never independent.

Example: one card from a pack; "king" and "heart" are independent since $\dfrac{1}{52} = \dfrac{1}{13} \times \dfrac{1}{4}$.

Events $E_1, \ldots, E_n$ form a partition of $S$ if $E_i \cap E_j = \varnothing$ for $i \ne j$, $E_1 \cup \cdots \cup E_n = S$ and $P(E_i) > 0$ for every $i$.

Theorem of total probability: for any event $A$,

In a tree: multiply along every path ending in $A$ and add. Example: bag I ($5$ white, $3$ black) and bag II ($2$ white, $6$ black) chosen by a coin: $P(W) = \dfrac{7}{16}$.

If $E_1, \ldots, E_n$ partition $S$ and $P(A) > 0$,

• $P(E_i)$: prior; $P(A \mid E_i)$: likelihood; $P(E_i \mid A)$: posterior.
• The posteriors add to $1$.
• A rare cause can stay unlikely even after strong evidence: in a screening test with prior $5\%$, a positive result gives only $\dfrac{9}{28} \approx 0.32$.

Example: a complaint at a call centre with shifts $40\%$, $35\%$, $25\%$ and complaint rates $2\%$, $3\%$, $4\%$: $P(C) = 0.0285$; posteriors $\dfrac{16}{57}$, $\dfrac{7}{19}$, $\dfrac{20}{57}$.

1. Define every event in words.
2. Read the question: does it say given? Which event is known?
3. Events in order → multiplication rule (tree). Unrelated trials → independence.
4. Outcome through several causes → total probability. Cause after the outcome → Bayes.
5. Check: probabilities between $0$ and $1$; posteriors add to $1$; the answer makes sense in context.

Not in the 2026–27 syllabus: random variables and their probability distributions, mean and variance of a random variable, Bernoulli trials and the binomial distribution.