# Conditional Probability and Independence

Mathematics I · Data Analysis and Probability · https://tryals.app/learn/mathematics-i/conditional-probability-and-independence

## Conditional Probability

New information changes probabilities. The **conditional probability** of $A$ given $B$, written $P(A \mid B)$, is the probability that $A$ occurs knowing $B$ has occurred:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0$$

Conditioning **restricts the sample space** to outcomes where $B$ holds, asking what fraction of those also have $A$. This is how evidence updates belief.

Rearranging the formula gives the **multiplication rule**: 

$$P(A \cap B) = P(A \mid B)\,P(B)$$

This states the joint probability equals one event's probability times the other's conditional probability.

## Independence vs. Exclusion

Two events are **independent** when knowing one occurred does not change the other's probability: $P(A \mid B) = P(A)$, or symmetrically:

$$P(A \cap B) = P(A)\,P(B)$$

Independent probabilities multiply. Two fair coin flips landing heads have probability $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.

| Property | Mutually Exclusive | Independent |
| --- | --- | --- |
| **Definition** | Cannot co-occur | No effect on each other |
| **Equation** | $P(A \cap B) = 0$ | $P(A \cap B) = P(A)P(B)$ |
| **Relationship** | Strongly dependent | Unrelated probabilities |

> **Common pitfall:** Confusing **independent** with **mutually exclusive**. Mutually exclusive events are maximally dependent because if one happens, the other cannot.

## Practice questions

7 of this lesson's 13 practice questions, with answers. The full set is in the app.

### 1. Select every statement that is equivalent to 'A and B are independent', assuming both have positive probability.

A. P(A and B) = 0
B. P(A given B) = P(A)
C. P(A and B) = P(A) times P(B)
D. P(B given A) = P(B)
E. P(A or B) = P(A) + P(B)

**Answer:** B. P(A given B) = P(A); C. P(A and B) = P(A) times P(B); D. P(B given A) = P(B)

**Why:** The first three are the same statement written three ways, and each implies the others. The last two both describe disjoint events, which for positive probabilities is exactly the opposite situation.

Page: https://tryals.app/practice/mathematics-i/conditional-probability-and-independence/select-every-statement-that-is-equivalent-to-a-and-b-are

### 2. Events A and B satisfy P(A and B) = 0.12 and P(B) = 0.3. Compute P(A given B).

**Answer:** 0.4 (within ±0.005)

**Why:** P(A given B) = 0.12/0.3 = 0.4. The denominator is the _condition_, never the other event: dividing by P(A) instead answers a different question. Conditioning rescales the shrunken sample space back to a total of 1.

Page: https://tryals.app/practice/mathematics-i/conditional-probability-and-independence/events-a-and-b-satisfy-p-a-and-b-0-12-and-p-b-0-3-compute-p-a

### 3. Two mutually exclusive events with non-zero probability cannot be independent.

**Answer:** True

**Why:** True. If A occurs, B cannot, so $P(B \mid A) = 0 \neq P(B)$, the opposite of independence. The two terms sound similar and are nearly contradictory: exclusivity is a very strong *dependence*. Only a probability-zero event can be both.

Page: https://tryals.app/practice/mathematics-i/conditional-probability-and-independence/two-mutually-exclusive-events-with-non-zero-probability-cannot-be

### 4. Using that table of 100 people, compute the probability that a person is a student given that they like coffee.

**Answer:** 0.75 (within ±0.005)

**Why:** 30/40 = 0.75. Note that P(likes coffee given student) is 30/50 = 0.6 instead, the same joint cell over a different condition. Swapping the two is the most common conditional-probability error, and the table makes the difference visible.

Page: https://tryals.app/practice/mathematics-i/conditional-probability-and-independence/using-that-table-of-100-people-compute-the-probability-that-a-person

### 5. A and B are independent, with P(A) = 0.6 and P(B) = 0.5. Compute P(A and B).

**Answer:** 0.3 (within ±0.005)

**Why:** Independent probabilities multiply: 0.6 times 0.5 = 0.3. Note this is not zero, so A and B can perfectly well both occur, independent events are never mutually exclusive unless one of them is impossible.

Page: https://tryals.app/practice/mathematics-i/conditional-probability-and-independence/a-and-b-are-independent-with-p-a-0-6-and-p-b-0-5-compute-p-a

### 6. A test for a disease affecting 1 in 1000 people is 99% accurate. You test positive. Why is the chance you have it closer to 9% than to 99%?

A. Because accuracy and probability are unrelated quantities
B. Because false positives from 999 healthy people outnumber the 1 true one
C. Because 99% accuracy refers to negative results only
D. Because the test's accuracy figure is misleading and the true accuracy is lower

**Answer:** B. Because false positives from 999 healthy people outnumber the 1 true one

**Why:** In 1000 people: one true positive, and about ten false ones from the 999 healthy. So roughly one positive in eleven is real. Base rates dominate when a condition is rare, the reason screening programmes for rare diseases generate so much needless alarm, and the commonest failure of medical intuition.

Page: https://tryals.app/practice/mathematics-i/conditional-probability-and-independence/a-test-for-a-disease-affecting-1-in-1000-people-is-99-accurate-you

### 7. Complete the relationship between the two most-confused notions.

**Answer:** Two events with positive probability that are mutually exclusive must be **dependent**.

**Why:** If A occurs, B is ruled out entirely, so P(B given A) = 0, which is not P(B) unless B was impossible to begin with. Learning that one occurred changes the other's probability as much as it possibly could, that is extreme dependence, the opposite of independence.

Page: https://tryals.app/practice/mathematics-i/conditional-probability-and-independence/complete-the-relationship-between-the-two-most-confused-notions