Combined events and tree diagramsIB MYP Maths Extended: Revision notes

The probability of an event is a number from $0$ to $1$: $P(\text{event})=\frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}$. The probabilities of all possible outcomes add to $1$, so the complement (the event not happening) has probability $1-P(\text{event})$. Example: a bag holds $3$ red and $2$ green counters, so $P(\text{red})=\frac{3}{5}$ and $P(\text{not red})=1-\frac{3}{5}=\frac{2}{5}$.

Two events are independent if one happening does not change the probability of the other, for example two spins of a spinner or a coin toss and a die roll. For independent events A and B, the multiplication rule gives $P(A\text{ and }B)=P(A)\times P(B).$ Example: a spinner lands on green with probability $0.3$. Two spins are independent, so $P(\text{two greens})=0.3\times0.3=0.09$. Multiply when you see 'and', 'then' or 'both'.

Events are mutually exclusive if they cannot happen at the same time, such as rolling a $2$ and rolling a $5$ on one die throw. For mutually exclusive events the addition rule gives $P(A\text{ or }B)=P(A)+P(B).$ Example: for a fair die, $P(2\text{ or }5)=\frac16+\frac16=\frac13$. Add when you see 'or' for events that cannot both happen, and when you combine different routes that give the same overall outcome.

A tree diagram shows every outcome of a sequence of events. Each branch is labelled with its probability and the branches from one point add to $1$. To find the probability of a route, multiply along its branches. To find the probability of an outcome that can happen by several routes, add the route probabilities. Example: a spinner has $P(\text{green})=0.3$ and $P(\text{yellow})=0.7$, spun twice. Exactly one green: green then yellow is $0.3\times0.7=0.21$ and yellow then green is $0.7\times0.3=0.21$, so $P=0.21+0.21=0.42$. When items are replaced, the branch probabilities are the same on every draw, so the events are independent.

If an item is not replaced, the numbers change for the second draw, so the events are dependent and the second-branch probabilities depend on what happened first. Example: a drawer has $6$ black and $4$ white socks. Two are taken without replacement. $P(\text{both black})=\frac{6}{10}\times\frac{5}{9}=\frac{1}{3}$, because after one black sock is removed, $5$ black remain out of $9$. Different colours: $\frac{6}{10}\times\frac{4}{9}+\frac{4}{10}\times\frac{6}{9}=\frac{48}{90}=\frac{8}{15}$.

'At least one' is often quickest as $1-P(\text{none})$. Example: $P(\text{at least one yellow in two spins})=1-0.3\times0.3=0.91$. Method: (1) decide with or without replacement; (2) write each route; (3) multiply along routes; (4) add routes for the same outcome; (5) check that all outcomes add to $1$. Leave answers as fractions or decimals (3 s.f.), and show each step so a marker can follow your working.