Independent and Conditional Probability
- In probability, we often perform an experiment (rolling a die, drawing a ticket, checking whether a train is late).
- An event is a set of outcomes we care about, such as:
- “rolling a 6”
- “the train is late”
- “the student chooses trampolining”
- In this chapter we focus on:
- Conditional probability – probabilities given that we already know something happened.
- Independent events – situations where this extra information does not change the probability.
- These two ideas are closely linked.
Conditional Probability: “Given That”
Conditional probability
The conditional probability of an event $B$ given that $A$ has occurred is written $P(B \mid A)$. The conditional probability of an event $B$ given that $A$ has occurred is written $P(B \mid A)$.
When you read “given that”, you should think:
The sample space has been restricted to the outcomes where $A$ happens.
So, outside $A$ we no longer care about any outcomes. Inside $A$ we ask: what fraction also lie in $B$?
Conditional Probability Formula
Using this “reduced sample space” idea, we get the rule
$$P(B \mid A)=\frac{P(A \cap B)}{P(A)}, \quad P(A)>0 $$
- The denominator is $P(A)$ because we are working inside $A$.
- The numerator is $P(A \cap B)$: the overlap of $A$ and $B$.
Conditional Probability with Venn Diagrams and Tables
Venn Diagram View
- In a Venn diagram:
- The "new sample space" for $P(B \mid A)$ is the region $A$.
- The favourable outcomes are in the overlap $A \cap B$.
- So, using counts, $$P(B \mid A)=\frac{n(A \cap B)}{n(A)} .$$
- The probability version is the same idea with probabilities instead of counts.
Two-Way (Contingency) Tables
For real data, probabilities often come from a two-way table (for example, gender vs chosen sport).
To find something like $P \text { (Table tennis } \mid \text { Female) }$:
- Identify the condition (“Female”) → use the female row total as the denominator.
- Use the intersection cell (“Female and Table tennis”) as the numerator.
- Form the fraction and simplify.
This method is the same as the Venn diagram approach, just organised in a table.
Joint Probability and Tree Diagrams
Multiplication Rule (General Form)
- Rearranging the conditional probability formula gives $$P(A \cap B)=P(A) P(B \mid A)$$
- This is the joint probability rule: to find the probability that both $A$ and $B$ happen, multiply:
- the probability of the first event $A$, and
- the conditional probability of the second given the first, $P(B \mid A)$.
Tree Diagrams
A tree diagram organises multi-stage experiments (for example, “train late?” then “miss appointment?”).
- Rules:
- Multiply along a branch to get a joint probability, e.g. $P(A \cap B)=P(A) P(B \mid A)$.
- Add along different branches (mutually exclusive paths) to get the probability of an event.
- Tree diagrams are especially helpful when probabilities change after the first event, such as:
- problems without replacement
- “if it rains, then …” style problems
- Each second-stage probability is then a conditional probability.
Independent Events
Independent events
Events $A$ and $B$ are independent if knowing that $A$ happened does not change the probability of $B$. Formally, $P(B\mid A)=P(B)$.
Independence is a statement about information: if event $A$ occurs (or does not occur), it does not give you any extra help in predicting event $B$.
- Independence is like two separate light switches controlling two different lamps.
- Switching one on does not affect whether the other lamp is on.
Mutual Exclusivity Is Different From Independence
Students often confuse independent with mutually exclusive.
Mutually exclusive events
Two events are mutually exclusive if they cannot occur at the same time, so $A\cap B=\varnothing$.
- For a fair six-sided die, let $A$ be "roll a 1" and $B$ be "roll a 2".
- These events are mutually exclusive: you cannot roll both 1 and 2 on the same throw, so $P(A\cap B)=0$.
- Using the addition rule, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac16+\frac16-0=\frac{2}{6}$$
- But are $A$ and $B$ independent? If $A$ happens, then $B$ definitely does not happen.
- So knowing $A$ changes the probability of $B$ to 0.
- Therefore, mutually exclusive events (with non-zero probabilities) are not independent.
- Do not say "mutually exclusive" when you mean "independent".
- If $A$ and $B$ are mutually exclusive and both have positive probability, then $P(B\mid A)=0$, so they cannot be independent.
Three Equivalent Ways To Test Independence
- There are several equivalent ways to express independence.
- In practice, you choose the one that matches the information you are given.
Method 1: Multiplication Rule
- If $A$ and $B$ are independent, then $$P(A\cap B)=P(A)\times P(B)$$
- This is useful when you can find $P(A\cap B)$ directly (for example from a table, Venn diagram, or tree diagram).
Method 2: Conditional Probability Form
- If $A$ and $B$ are independent, then $$P(B\mid A)=P(B\mid A')=P(B)$$ where $A'$ means "not $A$".
- This matches the meaning of independence: the probability of $B$ is unchanged even after learning whether $A$ happened.
Method 3: Conditional Probability Formula
- Conditional probability is defined by $$P(B\mid A)=\frac{P(A\cap B)}{P(A)} \quad (P(A)>0)$$
- If events are independent, then $P(A\cap B)=P(A)P(B)$, and substituting into the formula gives $P(B\mid A)=P(B)$.
Conditional probability will be covered in more detail in the next article.
The three statements
- $P(A\cap B)=P(A)P(B)$,
- $P(B\mid A)=P(B)$ (when $P(A)>0$), and
- $P(A\mid B)=P(A)$ (when $P(B)>0$)
are all equivalent ways of saying "$A$ and $B$ are independent".
Tickets In A Bag (Venn Diagram Approach)
- Tickets numbered $1$ to $9$ are in a bag and one ticket is chosen at random.
- Let
- $A$: "the number is even"
- $B$: "the number is a square number".
- The sample space has 9 equally likely outcomes.
- Even numbers: $\{2,4,6,8\}$ so $P(A)=\frac{4}{9}$
- Square numbers: $\{1,4,9\}$ so $P(B)=\frac{3}{9}=\frac{1}{3}$
- Intersection: $A\cap B=\{4\}$ so $P(A\cap B)=\frac{1}{9}$
- To test independence with the multiplication rule: $$P(A)P(B)=\frac{4}{9}\times\frac{3}{9}=\frac{12}{81}=\frac{4}{27}$$
- But $$P(A\cap B)=\frac{1}{9}=\frac{3}{27}$$
- Since $\frac{1}{9}\ne \frac{4}{27}$, the events are not independent.
- You can also check with conditional probability:
- $$P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{9}}{\frac{4}{9}}=\frac14$$ whereas $P(B)=\frac13$.
- Because $\frac14\ne \frac13$, knowing the ticket is even changes the chance it is square.
Train Lateness And Missing An Appointment (Tree Diagram Approach)
- A train is late with probability $\frac14$.
- If it is late, Minnie misses her dentist appointment with probability $\frac35$.
- If it is not late, she misses with probability $\frac15$.
- Let
- $A$: "train is late"
- $B$: "Minnie misses appointment".
- From the tree diagram rule, $$P(A\cap B)=P(A)\times P(B\mid A)=\frac14\times\frac35=\frac{3}{20}$$
- Also, $$P(A'\cap B)=P(A')\times P(B\mid A')=\frac34\times\frac15=\frac{3}{20}$$
- So $$P(B)=P(A\cap B)+P(A'\cap B)=\frac{3}{20}+\frac{3}{20}=\frac{6}{20}=\frac{3}{10}$$
- Now test independence.
- Theorem 2 test: $$P(A)P(B)=\frac14\times\frac{3}{10}=\frac{3}{40} \ne \frac{3}{20}=P(A\cap B)$$ So $A$ and $B$ are not independent.
- Theorem 3 test: $$P(B\mid A)=\frac35, \quad P(B\mid A')=\frac15, \quad P(B)=\frac{3}{10}$$ These are not equal, so the events are dependent.
- When a question gives you "if ... then ..." information, it is usually giving conditional probabilities.
- Draw a tree diagram first, label $P(B\mid A)$ and $P(B\mid A')$, then multiply along branches.
Independence In Two-Way Tables (Reasoning With Data)
Independence also appears when working with data tables.
- Suppose students choose between trampolining and table tennis:
- Male: 39 trampolining, 16 table tennis (55 total)
- Female: 21 trampolining, 14 table tennis (35 total)
- Total students: $90$.
- Compute key probabilities:
- $P(\text{male and trampolining})=\frac{39}{90}$
- $P(\text{trampolining})=\frac{39+21}{90}=\frac{60}{90}=\frac23$
- $P(\text{male})=\frac{55}{90}$
- If gender and sport choice were independent, then $$P(\text{male and trampolining})=P(\text{male})P(\text{trampolining})=\frac{55}{90}\cdot\frac{60}{90}$$
- Compare this product to $\frac{39}{90}$.
- If they are equal, the events are independent; if not, they are dependent.
With a two-way table, independence is often easiest to test by checking whether a conditional proportion matches the overall proportion, for example whether
$$P(\text{trampolining}\mid\text{male})\stackrel{?}{=}P(\text{trampolining}).$$
Common Patterns: When Independence Is Plausible (And When It Is Not)
- Independence is not something you assume automatically, you justify it.
- Events are often independent when:
- They come from separate experiments (coin flip and die roll).
- There is replacement (drawing a card, replacing it, then drawing again).
- The process is designed so one outcome cannot influence the next.
- Events are often dependent when:
- You sample without replacement (taking two doughnuts from a bag and eating the first).
- One event is a cause or strong influence of the other (rain affects playing tennis).
- The information about one event changes the sample space for the other.
- Be careful with "two draws" problems.
- If you do not replace the first item, probabilities on the second draw usually change, so events are dependent.
- Explain, in words, what it means for two events to be independent.
- For a fair die rolled twice, let $A$ be "first roll is 6" and $B$ be "second roll is 6". Are $A$ and $B$ independent? Justify using $P(A\cap B)=P(A)P(B)$.
- In the ticket example, compute $P(A\mid B)$ and compare it to $P(A)$. What does this tell you about independence?