Difficulty: Intermediate – Advanced | Prerequisites: Lessons 3 – 4 (probability rules, Venn diagrams, two-way tables).
Conditional probability asks: how does knowing one thing happened change the probability of another? This is one of the most important ideas in all of statistics. It is the concept behind medical test accuracy, courtroom evidence, spam filters, and virtually every inference procedure you will encounter later in the course. Independence is the special case where knowing one event tells you nothing about the other. Tree diagrams give you a visual method for computing joint and conditional probabilities when the problem has a sequential structure.
Conditional probability, P(A|B), is the probability of A given that B has occurred. You calculate it by dividing the joint probability P(A ∩ B) by P(B). Two events are independent if knowing one does not change the probability of the other, i.e. P(A|B) = P(A). Tree diagrams break a multi-stage process into branches, making it straightforward to find joint probabilities by multiplying along branches and to find overall probabilities by adding across branches.
Conditional probability
The probability that event A occurs given that event B has already occurred. Written P(A|B), read "the probability of A given B."
In simple terms, you are zooming in on just the cases where B happened, and asking how often A also happened within that subset.
Independence (of events)
Two events A and B are independent if the occurrence of one does not change the probability of the other. Formally: P(A|B) = P(A), or equivalently P(B|A) = P(B).
Think of it as: knowing B happened gives you no useful information about whether A happened.
General multiplication rule
P(A ∩ B) = P(B) × P(A|B). The probability that both events occur equals the probability of the first times the conditional probability of the second given the first.
In simple terms, multiply along the chain of "what has to happen first, then what has to happen next given the first thing happened."
Multiplication rule for independent events
If A and B are independent, P(A ∩ B) = P(A) × P(B). The conditional part drops out because P(A|B) = P(A).
Tree diagram
A branching diagram where each branch represents a stage of a random process. Branch probabilities are conditional probabilities. Multiplying along a complete path from root to leaf gives the joint probability of that sequence of outcomes.
In simple terms, it is a map of all the ways a multi-step process can unfold, with probabilities attached to each fork.
The most intuitive way to see conditional probability: restrict your attention to one row or one column, and find the proportion within that restricted group.
Using the Squid Games / Taco table:
Eats a Taco | Does Not Eat a Taco | Total | |
|---|---|---|---|
Watched Squid Games | 21 | 40 | 61 |
Did Not Watch | 29 | 10 | 39 |
Total | 50 | 50 | 100 |
P(S | E) = "given they ate a taco, what is the probability they watched Squid Games?" Look only at the 50 people who ate a taco. Of those, 21 watched Squid Games. So P(S | E) = 21/50 = 0.42.
P(E | S^c) = "given they did not watch Squid Games, what is the probability they ate a taco?" Look only at the 39 who did not watch. Of those, 29 ate a taco. So P(E | S^c) = 29/39 ≈ 0.744.
This is exactly what conditional relative frequency was in the exploratory data unit, now with formal probability notation.
P(A|B) = P(A ∩ B) / P(B)
The denominator is the probability of the "given" event. You are re-scaling the joint probability to account for the reduced sample space.
Example: P(Hit) = 0.7 for a missile. P(both missiles hit) = 0.57. P(second hits | first hits) = P(both hit) / P(first hits) = 0.57 / 0.7 ≈ 0.814.
Rearranging the conditional probability formula: P(A ∩ B) = P(B) × P(A|B).
This is on the AP formula sheet (the conditional probability formula at the top, which rearranges to this).
Example: P(Hit) = 0.7 for the first missile. If the first hits, P(second hits | first hits) = 0.95. P(both hit) = 0.7 × 0.95 = 0.665.
P(all three cards are red, drawn without replacement from a 52-card deck):
P(1st red) = 26/52
P(2nd red | 1st red) = 25/51
P(3rd red | first two red) = 24/50
P(all three red) = (26/52) × (25/51) × (24/50) ≈ 0.1176
Events A and B are independent if P(A|B) = P(A). Equivalently, P(B|A) = P(B). Equivalently, P(A ∩ B) = P(A) × P(B).
To test independence from a two-way table, compute P(A|B) and compare it to P(A). If they are equal (or very close, in a sample), the events are independent. If they differ, the events are not independent.
Example (gender and dominant hand): P(left-handed) = 10/100 = 0.10. P(left-handed | male) = 7/46 ≈ 0.152. Since 0.152 ≠ 0.10, the events "male" and "left-handed" are not independent in this sample.
Mutually exclusive events cannot happen at the same time. P(A ∩ B) = 0.
If A and B are mutually exclusive and both have non-zero probability, they are definitely not independent. Here is why: if A has occurred, then B certainly has not, so P(B|A) = 0, which is not equal to P(B).
In short, mutually exclusive events with non-zero probabilities are as far from independent as possible, because the occurrence of one completely rules out the other.
When A and B are independent, P(A|B) = P(A), so the general multiplication rule simplifies to P(A ∩ B) = P(A) × P(B).
Example: roll a die and flip a coin. P(roll a 3 and flip heads) = (1/6) × (1/2) = 1/12 ≈ 0.083. The die and coin do not influence each other, so independence applies.
Draw the first stage as branches from a single starting point (e.g. "has cancer" / "does not have cancer").
From each first-stage branch, draw second-stage branches (e.g. "positive mammogram" / "negative mammogram").
Label each branch with its conditional probability.
Multiply along a path to get the joint probability of that sequence.
Add joint probabilities across relevant paths to find overall probabilities.
Given information:
P(cancer) = 0.01, so P(no cancer) = 0.99
P(negative mammogram | cancer) = 0.03, so P(positive | cancer) = 0.97
P(positive mammogram | no cancer) = 0.06
Tree structure:
Cancer (0.01) → Positive (0.97): joint = 0.01 × 0.97 = 0.0097
Cancer (0.01) → Negative (0.03): joint = 0.01 × 0.03 = 0.0003
No cancer (0.99) → Positive (0.06): joint = 0.99 × 0.06 = 0.0594
No cancer (0.99) → Negative (0.94): joint = 0.99 × 0.94 = 0.9306
P(positive mammogram) = 0.0097 + 0.0594 = 0.0691
P(cancer | positive) = P(cancer ∩ positive) / P(positive) = 0.0097 / 0.0691 ≈ 0.140
So even with a positive mammogram, the probability of actually having cancer is only about 14%. This is a well-known and counterintuitive result: when the disease is rare, most positive tests are false positives.
P(multiple birth) = 0.035, P(single birth) = 0.965
P(left-handed | multiple birth) = 0.22
P(left-handed | single birth) = 0.11
P(left-handed) = (0.035 × 0.22) + (0.965 × 0.11) = 0.0077 + 0.10615 = 0.11385
P(multiple birth | left-handed) = 0.0077 / 0.11385 ≈ 0.0677
About 6.8% of left-handed children come from multiple births, compared to the baseline 3.5% of all children. Being left-handed makes it roughly twice as likely (in relative terms) that the child is from a multiple birth.
Conditional probability formula (on the AP formula sheet):
P(A|B) = P(A ∩ B) / P(B)
General multiplication rule (derived from the above):
P(A ∩ B) = P(B) × P(A|B)
Multiplication rule for independent events:
P(A ∩ B) = P(A) × P(B) [only when A and B are independent]
Test for independence:
Check whether P(A|B) = P(A). If yes, independent. If no, not independent.
Medical diagnostics rely entirely on conditional probability. When a doctor says "the test is 97% accurate," they typically mean P(positive | disease), but what the patient wants to know is P(disease | positive), which can be dramatically different when the disease is rare. Spam filters use conditional probability (via Bayes' theorem, which is the tree-diagram logic formalised) to estimate P(spam | contains certain words). Insurance underwriting uses conditional probability to set different premiums based on risk factors.
Students often confuse P(A|B) with P(B|A). These are generally different quantities. P(cancer | positive test) is not the same as P(positive test | cancer). Getting these backwards is sometimes called the "prosecutor's fallacy."
Students assume that if two events are related, they are mutually exclusive, or vice versa. "Related" (not independent) and "cannot happen together" (mutually exclusive) are separate ideas.
Students forget to multiply along the branches of a tree diagram and instead try to add branch probabilities. You multiply along a single path (joint probability), then add across paths if you need a combined probability.
Students sometimes use the simplified multiplication rule P(A) × P(B) when events are not independent. Always check independence before using that shortcut. If there is no reason to believe the events are independent, use the general multiplication rule.
⚠️ The conditional probability formula P(A|B) = P(A ∩ B) / P(B) is on the AP formula sheet. Know how to use it in both directions: finding conditional probability from joint, and finding joint probability from conditional.
⚠️ The AP exam regularly asks "are events A and B independent? Justify your answer." You must show a numerical comparison, e.g. P(A|B) vs P(A), and state whether they are equal.
⚠️ Tree diagram questions appear on the free response section. You need to label every branch with the correct conditional probability, show the multiplication, and clearly state which paths you are adding.
⚠️ The mammogram-style problem (rare condition, imperfect test) is a classic AP exam question. The answer is always surprisingly low for P(condition | positive test). If your answer seems counterintuitively low, you are probably correct.
⚠️ Remember: mutually exclusive events with non-zero probabilities are not independent. This is tested in multiple-choice questions almost every year.
True or false: P(A|B) is always equal to P(B|A).
Fill in the blank: If two events are independent, then P(A ∩ B) = __________.
True or false: If A and B are mutually exclusive and both have positive probability, they are independent.
Fill in the blank: In a tree diagram, you __________ along a single path to get the joint probability, and you __________ across paths to get the total probability of an outcome.
True or false: P(disease | positive test) is usually much lower than P(positive test | disease) when the disease is rare.
Answers: 1. False (they are generally different). 2. P(A) × P(B). 3. False (they are definitely not independent). 4. Multiply; add. 5. True.
Q: Using the Squid Games / Taco table, find P(S | E), the probability someone watched Squid Games given that they ate a taco.
A: P(S | E) = P(S ∩ E) / P(E) = (21/100) / (50/100) = 21/50 = 0.42.
Q: A heat-seeking missile has P(Hit) = 0.7. If the first missile hits, the probability the second also hits is 0.95. What is the probability both missiles hit?
A: P(both hit) = P(first hits) × P(second hits | first hits) = 0.7 × 0.95 = 0.665.
Q: You draw three cards without replacement from a standard deck. What is the probability all three are red?
A: P(all red) = (26/52) × (25/51) × (24/50) = 15,600/132,600 ≈ 0.1176.
Q: Are "male" and "left-handed" independent in the Australian high school sample (P(left) = 0.10, P(left | male) ≈ 0.152)?
A: No. P(left-handed | male) ≈ 0.152, which does not equal P(left-handed) = 0.10. Since the conditional probability differs from the unconditional, the events are not independent.
Q: In the mammogram example, explain in one sentence why P(cancer | positive) is so much lower than P(positive | cancer).
A: Because cancer is very rare (1%), the large number of false positives from the 99% without cancer overwhelms the small number of true positives, so most women with a positive mammogram do not have cancer.
Conditional probability is the engine behind significance testing (Units 8 – 9). A p-value is a conditional probability: the probability of getting a result as extreme as the observed one, given that the null hypothesis is true. Independence comes back in the context of sampling: observations need to be independent for many inference procedures to be valid, and the 10% condition is a rule of thumb for approximately achieving that.
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