Conditional probability and independence

IB Mathematics: Applications and Interpretation SL· 4.8 Conditional probability and independence· 15 min read

1. Formal Definition of Conditional Probability★★☆☆☆⏱ 5 min

📘 Definition

Conditional Probability

P(A|B)

For any two events $A$ and $B$ where $P (B) > 0$ , the conditional probability of $A$ given $B$ is defined as:

Example:

Restricts the sample space to only outcomes where $B$ has occurred

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

📐 Worked Example

Given $P (A) = 0.6$ , $P (B) = 0.5$ , and $P (A \cap B) = 0.3$ , calculate $P (A ∣ B)$ .

1
1. Recall the conditional probability formula:
2
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
3
1. Substitute the given values and simplify:
4
$P(A|B) = \frac{0.3}{0.5} = 0.6$

2. Independent Events★★☆☆☆⏱ 5 min

📘 Definition

Independent Events

Two events $A$ and $B$ are independent if the occurrence of one does not change the probability of the other. This gives the core test condition for independence:

Example:

Independence does not mean no overlap, unlike mutually exclusive events

P(A \cap B) = P(A) \times P(B)

📐 Worked Example

Given $P (A) = 0.4$ , $P (B) = 0.5$ , and $P (A \cup B) = 0.7$ , test whether $A$ and $B$ are independent.

1
1. First calculate $P (A \cap B)$ using the addition rule for probability:
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$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = 0.2$
3
1. Calculate the product of the individual probabilities:
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$P(A) \times P(B) = 0.4 \times 0.5 = 0.2$
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1. Compare the two values: since $P (A \cap B) = P (A) \times P (B)$ , $A$ and $B$ are independent.

✓ Quick check

Test your understanding:

Which of the following is the correct test for independence of events $A$ and $B$ ?
- A: $P (A \cap B) = P (A) + P (B)$
- B: $P (A \cap B) = P (A) P (B)$
- C: $P (A ∣ B) = P (B)$
- D: $P (A \cup B) = P (A) P (B)$
Reveal answer
B —
The definition of independence gives this core condition, which is the most common test used in IB exams.

3. Conditional Probability with Diagrams★★★☆☆⏱ 8 min

✓ Calculator OK

Most exam problems for IB AI SL give probability information in Venn or tree diagrams. You can use these diagrams to calculate conditional probability directly without relying only on the formula.

📐 Worked Example

A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. Find the probability that the second marble is red, given the first marble drawn was red.

1
1. We know the first marble is red, so we restrict our sample space to this case. After drawing one red, 4 red and 3 blue marbles remain, for a total of 7 marbles.
2
1. The conditional probability is the number of remaining red marbles divided by total remaining marbles:
3
$P(R_2|R_1) = \frac{4}{7} \approx 0.571$
4
1. Verify with the formal formula to confirm:
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$P(R_2|R_1) = \frac{P(R_1 \cap R_2)}{P(R_1)} = \frac{(\frac{5}{8} \times \frac{4}{7})}{\frac{5}{8}} = \frac{4}{7}$

4. Common Pitfalls

Wrong move:

Swapping the numerator and denominator in the conditional formula, using $P (A ∣ B) = \frac{P ( A \cap B )}{P ( A )}$

Why:

Confusion about which event is the "given" event, leading to dividing by the wrong probability

Correct move:

The denominator is always the probability of the event after the $∣$ symbol. For $P (A ∣ B)$ , divide by $P (B)$ .

Wrong move:

Confusing mutually exclusive events with independent events

Why:

Assuming that no overlap means independence, which is incorrect

Correct move:

Use the $P (A \cap B) = P (A) P (B)$ test for independence, regardless of whether events are mutually exclusive.

Wrong move:

Treating draws without replacement as independent events

Why:

Assuming all random draws are automatically independent

Correct move:

For sampling without replacement, adjust probabilities for subsequent draws, as events are always dependent.

Wrong move:

Relying on context intuition to declare independence

Why:

Assuming that related events must be dependent, or unrelated events must be independent

Correct move:

Always apply the $P (A \cap B) = P (A) P (B)$ test to confirm independence, do not rely on context alone.

5. Quick Reference Cheatsheet

Concept	Formula/Condition	Key Note
Conditional Probability	$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) > 0$	Denominator = probability of the given event
Independence Test	$P (A \cap B) = P (A) \times P (B)$	Also equivalent to $P (A ∣ B) = P (A)$
Tree Diagram Conditional	Use the probability of the branch after the given event	No formula needed for simple cases
Venn Diagram Conditional	$\frac{∣ A \cap B ∣}{∣ B ∣}$	Works for counts and probabilities

6. Frequently Asked

What key phrases indicate I need to use conditional probability?

Phrases like given that, if, on the condition that all indicate you are working with a restricted sample space and need to calculate conditional probability.

Can two mutually exclusive events be independent?

Almost never. If two events are mutually exclusive, $P (A \cap B) = 0$ , which only equals $P (A) P (B)$ if at least one of $P (A)$ or $P (B)$ is 0, which is a trivial case.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Tree diagram conditional probability
2024 · 2
Test for event independence
2023 · 1
Venn diagram conditional probability

Going deeper

syllabusOfficial IB AI SL SyllabusSyllabus reference for probability

What's Next

Conditional probability and independence are core foundational concepts for all further probability and statistics study, and are heavily tested across both Paper 1 and Paper 2 of IB AI SL. You will encounter these concepts again when working with probability distributions, hypothesis testing, and statistical inference. Mastering the formal test for independence and correct application of the conditional probability formula will help you avoid common exam errors and earn full marks on calculation questions. Next, you can extend your knowledge with more complex probability problems and related topics.