Apply the rules of probability (addition, multiplication, conditional), permutations and combinations to calculate probabilities of compound events

What this dot point is asking

QCAA wants you to apply the rules of probability and counting techniques to find probabilities of compound events.

Probability rules

$P(A) =$ favourable outcomes / total outcomes (for equally likely outcomes).

$P(A^c) = 1 - P(A)$.

Addition rule. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. For mutually exclusive, $P(A \cap B) = 0$.

Multiplication rule. $P(A \cap B) = P(A) \cdot P(B | A)$. For independent, $P(B | A) = P(B)$ and $P(A \cap B) = P(A) P(B)$.

Conditional probability. $P(B | A) = P(A \cap B)/P(A)$.

Counting

Multiplication principle. $n$ ways for task $1$ and $m$ ways for task $2$ gives $nm$ ways combined.

Permutations (order matters). $^nP_r = \dfrac{n!}{(n-r)!}$.

Combinations (order does not matter). $^nC_r = \dfrac{n!}{r! (n-r)!}$.

With repetition. $n^r$ ways to choose $r$ from $n$ with replacement.

Choosing the right tool

Worked example

A bag holds $4$ red and $6$ blue marbles. Two are drawn without replacement. Find $P(\text{both red})$.

$P(\text{red on first}) = 4/10$. $P(\text{red on second} | \text{red on first}) = 3/9$.

$P(\text{both red}) = (4/10)(3/9) = 12/90 = 2/15$.

Alternatively: $^4C_2 / {}^{10}C_2 = 6/45 = 2/15$.

Common traps

Using $^nP_r$ when order doesn't matter. Selections are combinations.

Forgetting that without replacement reduces population. Second draw is from $n - 1$.

Treating dependent events as independent. Conditional probability is needed.

Double counting in addition. $P(A) + P(B)$ counts overlap twice.

In one sentence

Probability rules (addition $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, multiplication $P(A \cap B) = P(A) P(B|A)$, conditional $P(B|A) = P(A \cap B)/P(A)$) combine with counting principles (multiplication, permutations $^nP_r = n!/(n-r)!$ for ordered selections, combinations $^nC_r$ for unordered) to compute probabilities of compound events.