Probability

Basic probability

For equally likely outcomes:

$$P(E) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}, \qquad 0 \leq P(E) \leq 1$$

Complementary event: $P(E') = 1 - P(E)$.

Addition law

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

If $A$ and $B$ are mutually exclusive (cannot both happen), $P(A \cup B) = P(A) + P(B)$.

Multiplication law

$$P(A \cap B) = P(A) \times P(B \mid A)$$

If $A$ and $B$ are independent (one does not affect the other), $P(A \cap B) = P(A) \times P(B)$.

With and without replacement

Drawing 2 balls from 4 red and 6 blue:

• With replacement: $P(\text{both red}) = \frac{4}{10} \times \frac{4}{10}$
• Without replacement: $P(\text{both red}) = \frac{4}{10} \times \frac{3}{9} = \frac{2}{15}$

A tree diagram helps: multiply along branches, add between branches.

Using combinations

$$P(\text{exactly 2 girls when choosing 3 from 5 boys and 4 girls}) = \frac{C^4_2 \times C^5_1}{C^9_3}$$

Expected value

$$E = \sum (\text{value} \times \text{probability})$$

A game is fair if the expected gain is 0.

Common mistake: "at least one" problems. $P(\text{at least one}) = 1 - P(\text{none})$ is almost always the quickest method.