Probability

For an experiment with equally likely outcomes:

$P(A) = \frac{\text{number of outcomes that satisfy } A}{\text{total number of outcomes}}.$

$P$ ranges from $0$ (impossible) to $1$ (certain). Often written as a fraction, decimal, or percentage.

Worked examples (single die):

• $P(\text{roll a 4}) = \dfrac{1}{6}$.
• $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$ (outcomes 2, 4, 6).
• $P(\text{more than 4}) = \dfrac{2}{6} = \dfrac{1}{3}$ (outcomes 5, 6).

Complementary events: the event 'NOT A' is the complement of $A$, written $A'$: $P(A') = 1 - P(A).$

Useful when 'NOT A' is easier to count than 'A'.

Worked example. Probability of rolling AT LEAST one 6 in two rolls of a die.

• 'At least one 6' = 'not (no 6s)'.
• $P(\text{no 6 in one roll}) = 5/6$.
• $P(\text{no 6 in both rolls}) = (5/6)^2 = 25/36$.
• $P(\text{at least one 6}) = 1 - 25/36 = 11/36$.

'AND' (both events happen) for INDEPENDENT events: $P(A \text{ and } B) = P(A) \times P(B).$

Independent = the outcome of one doesn't affect the other (e.g. two separate die rolls).

Worked example. $P(\text{6 on first roll AND 4 on second roll}) = \dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$.

'OR' for MUTUALLY EXCLUSIVE events: $P(A \text{ or } B) = P(A) + P(B).$

Mutually exclusive = both cannot happen at once (e.g. roll a 2 OR roll a 5 — can't be both).

Worked example. $P(\text{roll 2 or 5}) = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}$.

General 'OR' rule (for ANY two events): $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

Subtract the overlap if events can BOTH happen.

Worked example. Pick a card. $P(\text{red or face card})$?

• $P(\text{red}) = 26/52$.
• $P(\text{face}) = 12/52$ (J, Q, K each in 4 suits).
• $P(\text{red AND face}) = 6/52$.
• $P(\text{red or face}) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13$.

A tree diagram lists every possible combined outcome of a multi-step experiment. Each branch represents a step; multiply along a branch to get its probability; add probabilities of branches that lead to the same outcome.

Worked example. A bag has 3 red and 5 blue balls. Two are drawn WITHOUT replacement. Find the probability of two red.

Tree:

• 1st draw: $P(\text{red}_1) = 3/8$.
• After taking a red, $P(\text{red}_2) = 2/7$ (one less red, one less total).

$P(\text{two reds}) = \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}.$

Independent vs dependent.

• WITH replacement: probabilities don't change.
• WITHOUT replacement: probabilities CHANGE on the second step.

Tree diagrams are ESPECIALLY useful for problems with conditional probabilities.