Probability Notes
Probability: Module 4 - Section 1 to 3 Overview
Basic Probability Concepts
Sample Space (S): The set of all possible outcomes of a random phenomenon.
Example: For devices (laptop, cellphone, tablets, calculator), if we consider the number of devices, the sample space could be S = {1, 2, 3, 4}, representing the number of devices a student uses.
Event: A subset of the sample space.
Example (Devices):
Let A = student uses 3 or more devices. If we assume the probabilities for number of devices are: P(1) = 0.3, P(2) = 0.4, P(3) = 0.2, P(4) = 0.1 (Total = 1).
Then A = {3, 4}. P(A) = P(3) + P(4) = 0.2 + 0.1 = 0.3.
Let B = student uses 2 or 3 devices.
B = {2, 3}. P(B) = P(2) + P(3) = 0.4 + 0.2 = 0.6.
Complement of A (A^c): All outcomes in S that are not in A.
A^c = {1, 2}. P(A^c) = P(1) + P(2) = 0.3 + 0.4 = 0.7. Alternatively, P(A^c) = 1 - P(A) = 1 - 0.3 = 0.7.
Intersection of A and B (A \cap B) (A and B): Outcomes common to both A and B.
A = {3, 4}, B = {2, 3}. So, A \cap B = {3}. P(A \cap B) = P(3) = 0.2.
Union of A and B (A \cup B) (A or B or both): Outcomes in A, in B, or in both.
A \cup B = {2, 3, 4}. P(A \cup B) = P(2) + P(3) + P(4) = 0.4 + 0.2 + 0.1 = 0.7.
Alternatively, using the Addition Rule: P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.6 - 0.2 = 0.7.
Venn Diagram: A visual representation used to understand probabilities, especially for unions and intersections of events.
Disjoint (Mutually Exclusive) Events
Definition: Two events are disjoint (or mutually exclusive) if they have no common outcomes. This means their intersection is an empty set \emptyset.
Example: Let C = student uses 1 device.
C = {1}. P(C) = 0.3.
A = {3, 4}. A \cap C = \emptyset (no common outcomes). Therefore, A and C are disjoint.
P(A \cap C) = 0.
For disjoint events, the Addition Rule simplifies: P(A \cup C) = P(A) + P(C) = 0.3 + 0.3 = 0.6.
B = {2, 3}. B \cap C = \emptyset. Therefore, B and C are disjoint.
P(B \cap C) = 0.
P(B \cup C) = P(B) + P(C) = 0.6 + 0.3 = 0.9.
Important Note: Disjoint events are NOT necessarily independent events.
Conditional Probability
Definition: The conditional probability of event A given event B is the probability that event A occurs, knowing that event B has already occurred.
Formula: If A and B are events with P(B) > 0, the conditional probability of A given B is defined by:
P(A | B) = \frac{P(A \cap B)}{P(B)}
Notes:
The intersection is commutative: P(A \cap B) = P(B \cap A).
Conditional probabilities are generally not commutative: P(A | B) \neq P(B | A).
Example (continued from devices):
What is the probability that a student uses 2 or 3 devices given they use 3 or more devices?
Find P(B | A), where B = uses 2 or 3 devices, A = uses 3 or more devices.
Recall P(A \cap B) = 0.2 and P(A) = 0.3.
P(B | A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.3} = \frac{2}{3} \approx 0.6667.
What is the probability that a student uses 1 device given they use 3 or more devices?
Find P(C | A), where C = uses 1 device, A = uses 3 or more devices.
Recall A \cap C = \emptyset, so P(A \cap C) = 0.
P(C | A) = \frac{P(A \cap C)}{P(A)} = \frac{0}{0.3} = 0. This makes sense, if a student uses 3 or more devices, they cannot also use just 1 device.
Independent and Dependent Events
We assume P(A) > 0 and P(B) > 0 for the following conditions.
Independent Events: Events A and B are independent if any one of these conditions is satisfied:
P(A | B) = P(A) (The occurrence of B does not change the probability of A).
P(B | A) = P(B) (The occurrence of A does not change the probability of B).
P(A \cap B) = P(A) P(B) (The Multiplication Rule for Independent Events).
Dependent Events: Events A and B are dependent if any one of these conditions is satisfied:
P(A | B) \neq P(A).
P(B | A) \neq P(B).
P(A \cap B) \neq P(A) P(B) (Note: The general multiplication rule P(A \cap B) = P(A)P(B|A) = P(A|B)P(B) always holds, regardless of independence).
Summary Table: Disjoint, Independent, and Dependent Events
Condition | Disjoint Events (Mutually Exclusive) | Independent Events | Dependent Events |
|---|---|---|---|
Intersection (A \cap B) | P(A \cap B) = 0 | P(A \cap B) = P(A)P(B) | P(A \cap B) \neq P(A)P(B) |
**Conditional Probability (P(A | B))** | P(A | B) = 0 (if P(B) > 0) |
Union (A \cup B) | P(A \cup B) = P(A) + P(B) | P(A \cup B) = P(A) + P(B) - P(A)P(B) | P(A \cup B) = P(A) + P(B) - P(A \cap B) |
Relationship between events | Cannot occur at the same time. Having one excludes the other. | Occurrence of one does not affect the probability of the other. | Occurrence of one affects the probability of the other. |
Further Examples and Applications
Example: Two Randomly Selected Students (using device probabilities)
Assume responses from two students are independent.
Recall: A = student uses 3 or more devices (P(A) = 0.3).
What is the probability that both use 3 or more devices?
Let A1 be the event for student 1 and A2 for student 2.
P(A1 \cap A2) = P(A1) P(A2) (due to independence) = 0.3 \times 0.3 = 0.09.
What is the probability that none of them uses 3 or more devices?
This means both use less than 3 devices, which is A^c.
Recall P(A^c) = 0.7.
P(A1^c \cap A2^c) = P(A1^c)P(A2^c) = 0.7 \times 0.7 = 0.49.
What is the probability that at least one of the 2 students uses 3 or more devices?
P(A1 \cup A2) = 1 - P((A1 \cup A2)^c) = 1 - P(A1^c \cap A2^c) = 1 - 0.49 = 0.51.
Alternatively: P(A1 \cup A2) = P(A1) + P(A2) - P(A1 \cap A2) = 0.3 + 0.3 - 0.09 = 0.51.
What is the probability that one uses one device and the other uses two devices?
Let D1 = uses 1 device (P(D1) = 0.3)
Let D2 = uses 2 devices (P(D2) = 0.4)
This can occur in two ways: (Student 1 uses 1 device AND Student 2 uses 2 devices) OR (Student 1 uses 2 devices AND Student 2 uses 1 device).
P((D{1,1} \cap D{2,2}) \cup (D{1,2} \cap D{2,1})) where D_{i,j} means student i uses j devices.
Since these two scenarios are disjoint and student responses are independent:
P(D{1,1} \cap D{2,2}) = P(D{1,1})P(D{2,2}) = 0.3 \times 0.4 = 0.12.
P(D{1,2} \cap D{2,1}) = P(D{1,2})P(D{2,1}) = 0.4 \times 0.3 = 0.12.
Total probability = 0.12 + 0.12 = 0.24.
Example: Pop Quiz with 3 Multiple Choice Questions
A quiz has 3 MC questions, each with 5 options.
Student guesses randomly, so P(C) = 1/5 = 0.2 (Correct) and P(I) = 4/5 = 0.8 (Incorrect).
Responses on each question are independent.
Possible outcomes (CCC, CCI, CIC, CII, ICC, ICI, IIC, III).
What is the probability that a student answers all questions correctly?
P({CCC}) = P(C)P(C)P(C) = 0.2 \times 0.2 \times 0.2 = 0.008.
What is the probability that a student answers one question correctly?
This means exactly one correct, two incorrect (e.g., IIC, ICI, CII).
P(IIC) = P(I)P(I)P(C) = 0.8 \times 0.8 \times 0.2 = 0.128.
There are 3 such combinations (C1 imes I2 imes I3, I1 imes C2 imes I3, I1 imes I2 imes C_3).
P(\text{one question correctly}) = 3 \times 0.128 = 0.384.
What is the probability that a student answers at least 1 question correctly?
P(\text{at least 1 correct}) = 1 - P(\text{none correct}).
P(\text{none correct}) = P({III}) = P(I)P(I)P(I) = 0.8 \times 0.8 \times 0.8 = 0.512.
P(\text{at least 1 correct}) = 1 - 0.512 = 0.488.
What is the probability that a student answers at least 2 questions correctly?
This means exactly 2 correct (CCI, CIC, ICC) or all 3 correct (CCC).
P(CCI) = P(C)P(C)P(I) = 0.2 \times 0.2 \times 0.8 = 0.032.
There are 3 such combinations for exactly 2 correct.
P(\text{at least 2 correct}) = P(CCC) + 3 \times P(CCI) = 0.008 + 3 \times 0.032 = 0.008 + 0.096 = 0.104.
Alternatively: 1 - P(\text{none or 1 correct}) = 1 - P(III) - P(\text{one question correct}) = 1 - 0.512 - 0.384 = 0.104.
Example: System Reliability
A system has three independent components: A, B, and C.
The signal passes if