Probability Applications

Definition. When all outcomes are equally likely: $P(A) = \dfrac{n(A)}{n(\xi)}.$

Range. $0 \le P(A) \le 1$. $P = 0$ means impossible; $P = 1$ means certain.

Forms. Cambridge accepts fractions, decimals, or percentages. Don't mix notations within one answer.

Worked. Bag with $5$ red, $3$ blue, $2$ green balls. One drawn at random.

• $P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2}$.
• $P(\text{blue or green}) = \dfrac{5}{10} = \dfrac{1}{2}$.
• $P(\text{yellow}) = 0$ (impossible).
• $P(\text{red, blue, or green}) = 1$ (certain).

Worked. Two fair dice rolled. Probability of total being $7$.

• Sample space: $36$ ordered pairs.
• Favourable: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — $6$ outcomes.
• $P = \dfrac{6}{36} = \dfrac{1}{6}$.

Complement rule. $A'$ is the event "$A$ does not happen". $P(A) + P(A') = 1, \quad P(A') = 1 - P(A).$

Use it when. "NOT $A$" is much easier to count than $A$ itself.

Worked. A bag has $4$ red and $6$ blue balls. Two are drawn without replacement. Probability of "at least one red".

• Easier: compute $P(\text{no red}) = P(\text{both blue})$.
• $P(\text{both blue}) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$.
• $P(\text{at least one red}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$.

Tip. "At least one" almost always begs for the complement: $1 - P(\text{none})$.

Addition rule (mutually exclusive events). If $A$ and $B$ cannot both happen at once: $P(A \cup B) = P(A) + P(B).$

General addition rule (when they CAN overlap): $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

Multiplication rule (independent events). If $A$ and $B$ are independent (one doesn't affect the other): $P(A \cap B) = P(A) \times P(B).$

Worked (mutually exclusive). Roll a die. $P(\text{prime}) = ?$

• Primes on a die: $2, 3, 5$ — three outcomes.
• $P = \dfrac{3}{6} = \dfrac{1}{2}$.

Worked (independent). Roll a die and toss a coin. $P(\text{6 AND heads})$.

• $P(6) = \dfrac{1}{6}$, $P(\text{H}) = \dfrac{1}{2}$.
• Independent: $P = \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$.

Worked (overlapping). Pick a card from a standard deck. $P(\text{red OR ace}).$

• $P(\text{red}) = \dfrac{26}{52}$, $P(\text{ace}) = \dfrac{4}{52}$, $P(\text{red ace}) = \dfrac{2}{52}$.
• $P = \dfrac{26 + 4 - 2}{52} = \dfrac{28}{52} = \dfrac{7}{13}$.

Relative frequency is an experimental estimate of probability when outcomes aren't all equally likely (a biased die, a biased coin): $\text{relative frequency} = \dfrac{\text{number of times event occurred}}{\text{total trials}}.$

As trials $\to \infty$, relative frequency converges to the true probability (the law of large numbers).

Worked. A spinner is spun $200$ times. Red appears $80$ times.

• Estimate of $P(\text{red}) = \dfrac{80}{200} = 0.4$.

Expected number. If an event has probability $P$ and we run $n$ trials, the expected count is: $E = P \times n.$

Worked. A die has $P(6) = \dfrac{1}{6}$. Roll it $300$ times. Expected number of sixes:

• $E = \dfrac{1}{6} \times 300 = 50$.

Tip. Cambridge often pairs "relative frequency" with "expected number" in a single question — use the relative frequency as the probability estimate, then multiply by $n$.