Combined probability

In probability, $P(A|B)$ is called conditional probability. We use it when we want the probability of one event, but we also have extra information about another event.

In $P(A|B)$, the bar "$|$" is read as “given that.” So we want the probability that $A$ occurs, but we are only looking at situations where $B$ is already true.

Option A is talking about the probability of $B$ happening when $A$ is the condition. That would be written as $P(B|A)$, not $P(A|B)$. So it does not match the meaning of $P(A|B)$.

Option B describes the probability that $A$ and $B$ happen at the same time in the original sample space. That is the intersection, written as $P(A\cap B)$. Conditional probability is different because it focuses on a restricted sample space where $B$ is already assumed to occur.

Option C suggests adding probabilities and dividing by total outcomes, which is not a standard definition for $P(A|B)$. Adding probabilities usually connects to “$A$ or $B$” ideas, not “$A$ given $B$.”

By definition, $P(A|B)$ means the probability that $A$ occurs when we already know $B$ has occurred.

Mathematically, the IB Data Booklet gives:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

This shows we take the overlap $A\cap B$ and compare it to all outcomes where $B$ happens (by dividing by $P(B)$). In other words, we are effectively limiting the sample space to outcomes where $B$ is true.

Option D correctly describes conditional probability: it finds the probability of $A$ after narrowing attention to the outcomes where $B$ has already happened.

In probability, $P(A|B)$ is conditional probability. We read it as “the probability of event $A$ given that event $B$ has occurred.” This idea is central in Topic 4: Sets, Logic and Probability.

A key formula from the IB Mathematics Data Booklet is:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

This tells us we’re measuring how likely $A$ is, but only within the outcomes where $B$ is already the case.

Option A describes finding the probability of $B$ under the condition that $A$ is known. That corresponds to $P(B|A)$. For $P(A|B)$, the event after the bar ($B$) is the one we are assuming has happened.

Option B is about both events happening together in the original sample space, which is the intersection $P(A\cap B)$. Although $P(A\cap B)$ is used in the conditional probability formula, it is not the definition of $P(A|B)$ itself.

Option C describes adding probabilities and dividing by total outcomes, which doesn’t match any standard rule that defines conditional probability. Conditional probability is about restricting the outcomes based on given information, not averaging sums like this.

After eliminating the incorrect choices, the correct description is the one that explains how conditional probability changes what we consider possible. For $P(A|B)$, we only look at outcomes where $B$ has happened, and then we find the probability that $A$ occurs within that restricted set.

The notation $P(A|B)$ means a conditional probability. We say it as “the probability that $A$ occurs given that $B$ has already occurred.”

A Venn diagram can help because it shows how knowing $B$ happened changes which outcomes we pay attention to.

Picture a Venn diagram with circles $A$ and $B$ inside a rectangle representing the universal set $U$. Normally, probabilities are measured using the entire rectangle.

But in $P(A|B)$, the “$|B$” tells us to treat $B$ as certain. Visually, that means we ignore everything outside circle $B$. The sample space effectively becomes just the region inside $B$.

Once circle $B$ is our new universe, we ask: within $B$, where does $A$ happen?

The only part of $A$ that matters now is the overlap with $B$, which is $A\cap B$. So $P(A|B)$ compares the overlap area to the entire area of $B$.

$\text{Conditioning = Restricting the sample space}$

Using the Venn diagram interpretation:

• Option A is $P(B|A)$ because it would make $A$ the restricted region.
• Option B is just the intersection $P(A\cap B)$ in the original full rectangle, not a conditional probability.
• Option C gives an operation that isn’t a standard probability definition.

The Venn diagram viewpoint matches the statement that we limit our attention to outcomes where $B$ has already occurred, and then we measure how likely $A$ is within that smaller set. That is exactly what conditional probability means.

The situation is a three-stage experiment (3 rounds). In a tree diagram, each stage is one level of branching. We are told that at each stage there are 3 possible outcomes (3 doors to choose from).

• Stage 1: 3 outcomes
• Stage 2: 3 outcomes
• Stage 3: 3 outcomes

To count the number of different complete paths in a multi-step process, we apply the Fundamental Counting Principle. It says that if there are $n$ ways for one step and $m$ ways for another step, then there are $n\times m$ ways to do both.

For three stages, we multiply the outcomes at each stage: $\text{Total Paths}={\text{Outcomes}}_1\times {\text{Outcomes}}_2\times {\text{Outcomes}}_3$

Replace each outcomes value with 3:

$\text{Total Paths}=3\times 3\times 3$

Now compute: $\begin{array}{rl}3\times 3& =9\\ 9\times 3& =27\end{array}$

So the tree diagram ends with $27$ different final branches.

Check the incorrect choices:

• Option A ($9$): This counts only two rounds ($3\times 3$).
• Option B ($12$): This could come from mixing operations (like adding) or using the wrong multiplication such as $3\times 4$.
• Option D ($6$): This is what you get if you add outcomes ($3+3$) instead of multiplying stages.

Since $3\times 3\times 3=27$, the completed tree diagram will have $27$ terminal nodes.

To count the number of terminal nodes in a tree diagram, focus on how many branches appear at each level. Here we know:

• The experiment has 3 stages.
• Each stage has 3 possible outcomes.

Because every stage branches the same way, the total number of endings follows an exponential pattern.

When each stage has the same number of outcomes, the total number of final outcomes is found using:

$\text{Total Outcomes}=(\text{outcomes per stage}{)}^{(\text{number of stages})}$

Each new stage multiplies the number of existing paths by the same factor, which connects to exponent rules.

Substitute 3 outcomes per stage and 3 stages:

$\begin{array}{rl}\text{Total Nodes}& =3^3\\ & =3\times 3\times 3\\ & =27\end{array}$

So there are 27 complete paths from start to finish.

Why the other choices are not correct:

• Option A ($9$): This matches $3^2$ (only two stages) or $3\times 3$.
• Option B ($12$): This doesn’t match the repeated-multiplication structure (it suggests something like $3\times 4$).
• Option D ($6$): This suggests adding ($3+3$) instead of multiplying branches.

Because $3^3=27$, the full tree diagram contains $27$ terminal nodes.