Conditional probabilityIB MYP Maths Extended: Revision notes

Conditional probability is the probability of an event given that another event has already happened. We write $P(A\mid B)$, read 'the probability of $A$ given $B$'. Knowing that $B$ has happened shrinks the sample space to the outcomes in $B$. The formula is $P(A\mid B)=\frac{P(A\text{ and }B)}{P(B)}.$ Example: $P(A\text{ and }B)=0.12$ and $P(B)=0.4$ give $P(A\mid B)=\frac{0.12}{0.4}=0.3$. Note that $P(A\mid B)$ and $P(B\mid A)$ are usually different.

A two-way table shows two features of the same group. To find a conditional probability, use only the row or column you are told about as the total. Example: Year 9 has 18 instrument players and 22 who do not play; Year 10 has 27 players and 13 non-players. Total players $=45$.

• $P(\text{instrument}\mid\text{Year 10})=\frac{27}{40}$ (the Year 10 total is 40).
• $P(\text{Year 9}\mid\text{instrument})=\frac{18}{45}=\frac25$ (the player total is 45). Check by the formula: $\frac{18/80}{45/80}=\frac{18}{45}$.

In a Venn diagram the circles show events and the overlap shows 'both'. Always fill in the overlap first, then the 'only' regions, then the outside. Example: 50 students; 30 study French ($F$), 22 study Spanish ($S$), 8 study both. French only $=30-8=22$, Spanish only $=22-8=14$, neither $=50-(22+8+14)=6$. For $P(S\mid F)$ use only the French circle (30 students), of whom 8 also study Spanish: $\frac{8}{30}=\frac{4}{15}$. For $P(F\mid S)$ use the Spanish circle (22 students): $\frac{8}{22}=\frac{4}{11}$.

Two events are dependent if the first outcome changes the probability of the second. This happens when you take items without replacement. Example: 6 ripe and 4 unripe mangoes; take two. After a ripe mango is removed, 5 ripe and 4 unripe remain among 9. So $P(\text{unripe second}\mid\text{ripe first})=\frac49$. On a tree diagram the second set of branches shows conditional probabilities. Multiply along branches for 'and' and add the branch products for 'or': $P(\text{both ripe})=\frac{6}{10}\times\frac59=\frac{30}{90}$, $P(\text{both unripe})=\frac{4}{10}\times\frac39=\frac{12}{90}$, so $P(\text{same})=\frac{42}{90}=\frac{7}{15}$. In general $P(A\text{ and }B)=P(A)\times P(B\mid A)$.

You can use conditional probability to test a claim: compare $P(A\mid B)$ with $P(A)$.

• If $P(A\mid B)>P(A)$, then $B$ makes $A$ more likely.
• If $P(A\mid B)<P(A)$, then $B$ makes $A$ less likely.
• If $P(A\mid B)=P(A)$, the events are independent. Example: $P(S)=\frac{22}{50}=0.44$ but $P(S\mid F)=\frac{4}{15}=0.267$, so studying French makes studying Spanish less likely. Write your conclusion in words about the context, and say what the limits of the data are.