# Introduction to Probability

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

## Quantifying Uncertainty

**Probability** measures how likely an event is, on a scale from 0 (impossible) to 1 (certain). It is the mathematics of uncertainty behind statistics and machine learning.

Key structural terms:
| Term | Definition & Example |
| --- | --- |
| **Sample Space** ($S$) | All possible outcomes (e.g., die roll: $\{1, 2, 3, 4, 5, 6\}$) |
| **Event** ($E$) | A subset of outcomes (e.g., rolling even: $\{2, 4, 6\}$) |

For **equally likely** outcomes, probability is a ratio:
$$P(E) = \frac{\text{favorable outcomes}}{\text{total outcomes}}$$
For a fair die, $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$.

## Rules and Pitfalls

Probability obeys **axioms**: $P(E) \ge 0$, $P(S) = 1$, and for **mutually exclusive** events (cannot both happen), probabilities **add**: $P(A \cup B) = P(A) + P(B)$.

From these come vital operational rules:
| Rule | Formula | Notes |
| --- | --- | --- |
| **Complement** | $P(\text{not } E) = 1 - P(E)$ | Best for "at least one" problems |
| **Addition** | $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | Prevents double-counting overlap |

> **Common pitfall:** Adding $P(A) + P(B)$ when events **can overlap**. Simple addition is only for **mutually exclusive** events; otherwise, you must subtract the intersection or risk exceeding 1.

## Practice questions

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

### 1. Complete the condition under which probabilities may simply be added.

**Answer:** P(A) + P(B) gives P(A or B) only when the two events are **mutually exclusive**.

**Why:** Only disjoint events can be added directly; otherwise their shared outcomes are counted twice. Independence is a different condition entirely, it licenses _multiplication_, not addition, and is nearly the opposite of mutual exclusivity.

Page: https://tryals.app/practice/mathematics-i/introduction-to-probability/complete-the-condition-under-which-probabilities-may-simply-be-added

### 2. The Monty Hall problem: three doors, you pick one, the host opens a different door revealing a goat and offers a switch. Switching wins two thirds of the time. Why does the host's action change the odds?

A. Because the host knows where the prize is, so his choice informs
B. Because the prize is moved after the initial choice
C. Because there are now two doors, so the probability is 1/2 either way
D. Because your original choice becomes invalid once a door is opened

**Answer:** A. Because the host knows where the prize is, so his choice informs

**Why:** Your door held 1/3 and still does; the other two held 2/3 between them, and the host, who cannot open the prize, has funnelled all of it onto one door. The hundred-door version makes it obvious, and the puzzle's fame comes from how firmly the wrong answer resists correction.

Page: https://tryals.app/practice/mathematics-i/introduction-to-probability/the-monty-hall-problem-three-doors-you-pick-one-the-host-opens-a

### 3. One card is drawn from a standard 52-card deck. Select every statement that is true.

A. P(heart) = 1/4
B. P(heart and king) = 0
C. P(heart or king) = 1/4 + 1/13
D. P(not a heart) = 3/4
E. P(king) = 1/13

**Answer:** A. P(heart) = 1/4; D. P(not a heart) = 3/4; E. P(king) = 1/13

**Why:** 13/52 = 1/4 and 4/52 = 1/13 are right, and the complement of a heart is 3/4. But the king of hearts is one card in both events, so P(heart and king) = 1/52, not 0, and P(heart or king) must subtract that overlap, giving 16/52.

Page: https://tryals.app/practice/mathematics-i/introduction-to-probability/one-card-is-drawn-from-a-standard-52-card-deck-select-every

### 4. If the probabilities of all outcomes sum to 1, no single outcome can have probability greater than 1.

**Answer:** True

**Why:** True. With every probability non-negative and the total exactly 1, no single term can exceed 1, one that did would force another negative. The two axioms together are what make probability a *measure*, and they are why a computed 'probability' above 1 is a sign the model is wrong.

Page: https://tryals.app/practice/mathematics-i/introduction-to-probability/if-the-probabilities-of-all-outcomes-sum-to-1-no-single-outcome-can

### 5. Two fair dice are rolled. How many of the 36 equally likely outcomes give a total of 7?

**Answer:** 6

**Why:** The pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), six of them, so the probability is 6/36 = 1/6. Seven is the most likely total precisely because it has the most ways to occur; counting outcomes is what makes that visible.

Page: https://tryals.app/practice/mathematics-i/introduction-to-probability/two-fair-dice-are-rolled-how-many-of-the-36-equally-likely-outcomes

### 6. A fair coin is flipped three times. Compute the probability of getting at least one head.

**Answer:** 0.875 (within ±0.005)

**Why:** P(no heads) = $(1/2)^{3}$ = 1/8, so P(at least one) = 1 - 1/8 = 7/8 = 0.875. 'At least one' problems are almost always easier through the complement, enumerating the seven favourable outcomes directly is far more work.

Page: https://tryals.app/practice/mathematics-i/introduction-to-probability/a-fair-coin-is-flipped-three-times-compute-the-probability-of

### 7. Events A and B satisfy P(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.2. Compute P(A or B).

**Answer:** 0.7 (within ±0.001)

**Why:** P(A or B) = 0.5 + 0.4 - 0.2 = 0.7. Adding without subtracting would give 0.9, counting the shared outcomes twice. This is inclusion-exclusion, the same principle already met with sets.

Page: https://tryals.app/practice/mathematics-i/introduction-to-probability/events-a-and-b-satisfy-p-a-0-5-p-b-0-4-and-p-a-and-b-0-2