2.1 Probability
2.1.1 Objectives
By the end of this unit, students will be able to:
- Explain the concept of randomness and how probability quantifies randomness.
- Recognize the basic concepts of sample space, equally likely outcomes, events, unions and intersections.
- Identify when two events are mutually exclusive, independent or complementary.
- Use probability rules to compute the probability of different types of events.
- Distinguish between marginal probability and conditional probability.
2.1.2 Overview
Probability provides a mathematical framework for describing and analyzing randomness. A random process (also called an experiment) is any activity or observation with uncertain outcomes.
Examples include:
- Rolling a six-sided die
- Tossing a coin twice
- Recording the temperature at a specific time
For each experiment, we can define the sample space (S), which is the set of all possible outcomes. For instance:
- A die roll has \(S = \{1,2,3,4,5,6\}\)
- Two coin tosses have \(S = \{HH, HT, TH, TT\}\)
Events and Set Operations
An event is any subset of the sample space. In probability, events behave like sets, so the following operations are essential:
- Union (A \(\cup\) B): outcomes in A or B or both
- Intersection (A \(\cap\) B): outcomes in both A and B
- Complement (\(A^c\)): outcomes in the sample space not in A
For example, if A = {even outcomes on a die} and B = {outcomes \(\ge\) 5}, then:
- A \(\cup\) B = {2,4,5,6}
- A \(\cap\) B = {6}
- \(A^c\) = {1,3,5}
Assigning Probabilities
If outcomes in a sample space are equally likely, then:
\[ P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes in S}} \]
For the die roll example:
\[ - P(A =even) = 3/6 = 1/2 - P(B=\ge 5) = 2/6 = 1/3 - P(A \cup B) = 4/6 = 2/3 \]
Probability Rules (Axioms)
From the general framework of probability:
- Total probability: \(P(S) = 1\)
- Boundedness: \(0 \leq P(A) \leq 1\) for any event A
- Addition rule: For any two events A and B,
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] - Special case – mutually exclusive events: If \(A \cap B = \emptyset\), then
\[ P(A \cup B) = P(A) + P(B) \] - Law of complements: \(P(A^c) = 1 - P(A)\)
Marginal, Joint, and Conditional Probability
When working with multiple events, we expand our probability framework:
- Marginal probability refers to the probability of a single event, ignoring others.
- Joint probability refers to the likelihood of two events happening together (A and B).
- Conditional probability is the probability of one event given another has occurred:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Image suggestion from PDF: Insert 2-way table of joint probabilities (Chapter 2, pg. 66–70).
Example
Suppose we toss a fair coin twice. The sample space is \(S = \{HH, HT, TH, TT\}\).
- Let A = “first toss is heads” = {HH, HT}
- Let B = “at least one tails” = {HT, TH, TT}
Then:
- \(P(A) = 2/4 = 0.5\)
- \(P(B) = 3/4 = 0.75\)
- \(P(A \cap B) = 1/4 = 0.25\)
- \(P(A \cup B) = 3/4 = 0.75\)
This illustrates how probability rules help us quantify uncertainty and reason about random outcomes.
2.1.4 Solved Exercises
Exercise 1
A set of 12 cards is numbered 1 through 12. A card is picked at random and the following events are defined:
- A: the number on the card is even
- B: the number on the card is greater than 8
Find:
- \(P(A)\)
- \(P(B)\)
- \(P(A \cap B)\)
- \(P(A \cup B)\)
Solution:
Since,
\(A = \{2,4,6,8,10,12\}\)
\(B = \{9,10,11,12\}\)
\(A \cap B = \{10,12\}\)
\(A \cup B = \{2,4,6,8,9,10,11,12\}\)
- \(P(A) = \tfrac{6}{12} = 0.5\)
- \(P(A) = \tfrac{6}{12} = 0.5\)
- \(P(B) = \tfrac{4}{12} = 0.333\)
- \(P(B) = \tfrac{4}{12} = 0.333\)
- \(P(A \cap B) = \tfrac{2}{12} = 0.167\)
- \(P(A \cap B) = \tfrac{2}{12} = 0.167\)
- \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.333 - 0.167 = 0.666\)
Exercise 2
A loaded spinner is divided into six regions labeled 1–6. The empirical probabilities are:
| Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Probability | 0.10 | 0.15 | 0.25 | 0.20 | 0.18 | 0.12 |
For events A (odd number) and B (number \(\le\) 3), compute:
- \(P(A)\)
- \(P(B)\)
- \(P(A \cup B)\)
- \(P(A \cap B)\)
- \(P(A^c)\)
- \(P(B^c)\)
Solution:
\(A = \{1,3,5\},\ B = \{1,2,3\}\)
\(A \cup B = \{1,2,3,5\},\ A \cap B = \{1,3\}\)
- \(P(A) = 0.10 + 0.25 + 0.18 = 0.53\)
- \(P(A) = 0.10 + 0.25 + 0.18 = 0.53\)
- \(P(B) = 0.10 + 0.15 + 0.25 = 0.50\)
- \(P(B) = 0.10 + 0.15 + 0.25 = 0.50\)
- \(P(A \cup B) = 0.10 + 0.15 + 0.25 + 0.18 = 0.68\)
- \(P(A \cup B) = 0.10 + 0.15 + 0.25 + 0.18 = 0.68\)
- \(P(A \cap B) = 0.10 + 0.25 = 0.35\)
- \(P(A \cap B) = 0.10 + 0.25 = 0.35\)
- \(P(A^c) = 1 - 0.53 = 0.47\)
- \(P(A^c) = 1 - 0.53 = 0.47\)
- \(P(B^c) = 1 - 0.50 = 0.50\)
Exercise 3
A group of 900 employees is classified by department (\(D_1\) = Sales, \(D_2\) = Tech) and by experience level (\(E_1\) = Entry, \(E_2\) = Mid, \(E_3\) = Senior).
| Entry (\(E_1\)) | Mid (\(E_2\)) | Senior (\(E_3\)) | Total | |
|---|---|---|---|---|
| Sales (\(D_1\)) | 180 | 150 | 120 | 450 |
| Tech (\(D_2\)) | 120 | 180 | 150 | 450 |
| Total | 300 | 330 | 270 | 900 |
Find the probability that a randomly selected employee:
- Is mid-level, \(P(E_2)\)
- Is a senior in Tech, \(P(D_2 \cap E_3)\)
- Works in Sales or is Senior, \(P(D_1 \cup E_3)\)
- Is not Entry-level, \(P(\text{not } E_1)\)
- Is not Tech and not Mid-level, \(P(\text{not } D_2 \cap \text{not } E_2)\)
Solution:
- \(P(E_2) = \tfrac{330}{900} = 0.367\)
- \(P(E_2) = \tfrac{330}{900} = 0.367\)
- \(P(D_2 \cap E_3) = \tfrac{150}{900} = 0.167\)
- \(P(D_2 \cap E_3) = \tfrac{150}{900} = 0.167\)
- \(P(D_1 \cup E_3) = P(D_1) + P(E_3) - P(D_1 \cap E_3)\)
\(= \tfrac{450}{900} + \tfrac{270}{900} - \tfrac{120}{900} = 0.667\)
- \(P(D_1 \cup E_3) = P(D_1) + P(E_3) - P(D_1 \cap E_3)\)
- \(P(\text{not } E_1) = 1 - \tfrac{300}{900} = 0.667\)
- \(P(\text{not } E_1) = 1 - \tfrac{300}{900} = 0.667\)
- \(P(\text{not } D_2 \cap \text{not } E_2) = \tfrac{180 + 120}{900} = \tfrac{300}{900} = 0.333\)
Exercise 4
Two fair coins are tossed — one red, one blue. The sample space has \(n=4\) equally likely outcomes:
\(\{(H,H), (H,T), (T,H), (T,T)\}\)
Let:
- \(A\): both show heads
- \(B\): exactly one head
- \(C\): at least one tail
Find:
- \(P(A)\)
- \(P(B)\)
- \(P(C)\)
- Which pairs of events are disjoint?
Solution:
- \(A = \{(H,H)\}\) → \(P(A) = \tfrac{1}{4}\)
- \(B = \{(H,T),(T,H)\}\) → \(P(B) = \tfrac{2}{4} = \tfrac{1}{2}\)
- \(C = \{(H,T),(T,H),(T,T)\}\) → \(P(C) = \tfrac{3}{4}\)
Disjoint pairs:
- \(A\) and \(B\) are disjoint
- \(A\) and \(C\) overlap (since \((H,H)\notin C\) so they’re actually disjoint too)
- \(B\) and \(C\) overlap (since \((H,T),(T,H) \in B \cap C\))
Exercise 5 (True/False)
If A and B are disjoint, then \(P(A \cup B) = P(A) + P(B)\).
Answer: TRUEFor any events A and B, \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\).
Answer: TRUE