Session 2 — Calculation of Probabilities
Decision Making Statistics — S04
This session develops the basic rules of probability and shows how to compute conditional probabilities in business situations.
Suggested pacing: Introduction (10 min), Basic calculation of probabilities (25 min), Detailed example (15 min), Application exercise (40 min).
1 Introduction
1.1 Random experiment, sample space, and events
A random experiment is an experiment whose outcome cannot be predicted with certainty in advance.
The sample space \(\Omega\) is the set of all possible outcomes of a random experiment.
An event is a subset of \(\Omega\). An elementary event contains only one outcome.
Example: when one customer file is selected at random,
- \(\Omega\) may be the set of all customer files,
- \(A\) may be the event “the customer is premium”,
- \(B\) may be the event “the customer bought online last month”.
2 Basic calculation of probabilities
2.1 Basic rules
For any event \(A\):
- \(P(A) \in [0,1]\)
- \(P(\Omega) = 1\)
- \(P(\emptyset) = 0\)
- \(P(\bar{A}) = 1 - P(A)\)
If all outcomes are equally likely, we use classical probability:
\[ P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{|A|}{|\Omega|} \]
2.2 Addition rule
For two events \(A\) and \(B\):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
If \(A\) and \(B\) are mutually exclusive, then:
\[ P(A \cup B) = P(A) + P(B) \]
2.3 Conditional probability and independence
Provided that \(P(B) > 0\),
\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]
Two events \(A\) and \(B\) are independent if:
\[ P(A \cap B) = P(A) \times P(B) \]
Equivalently, if \(P(B) > 0\),
\[ P(A \mid B) = P(A) \]
Do not confuse \(P(A \mid B)\) with \(P(B \mid A)\). These two probabilities are usually different.
2.4 Total probability formula
If \(B_1, B_2, \dots, B_n\) form a partition of \(\Omega\), then:
\[ P(A) = \sum_{i=1}^{n} P(A \mid B_i) \times P(B_i) \]
2.5 Bayes’ theorem
If \(B_1, B_2, \dots, B_n\) form a partition of \(\Omega\), then:
\[ P(B_k \mid A) = \frac{P(A \mid B_k) \times P(B_k)}{\sum_{i=1}^{n} P(A \mid B_i) \times P(B_i)} \]
When a probability question involves several sources, groups, or machines, draw a tree diagram or identify a partition before calculating.
3 Detailed example
A bank studies 200 clients:
- 120 use the mobile app,
- 80 have a premium account,
- 50 both use the app and have a premium account.
Let:
- \(A\) = “the client uses the mobile app”,
- \(B\) = “the client has a premium account”.
3.1 Union
\[ P(A) = \frac{120}{200} = 0.60, \qquad P(B) = \frac{80}{200} = 0.40, \qquad P(A \cap B) = \frac{50}{200} = 0.25 \]
Using the addition rule:
\[ P(A \cup B) = 0.60 + 0.40 - 0.25 = 0.75 \]
So the probability that a client uses the app or has a premium account is 0.75.
3.2 Conditional probability
The probability that a client has a premium account given that they use the app is:
\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.25}{0.60} \approx 0.417 \]
So about 41.7% of app users have a premium account.
4 Application exercise
ExerciseA factory has 3 machines (\(M_1\), \(M_2\), \(M_3\)) producing 40%, 35%, and 25% of output respectively. Defect rates are 2%, 3%, and 5%.
- What is the overall probability of picking a defective part?
- Given a defective part, what is the probability it came from \(M_3\)?
SolutionLet \(D\) be the event “the part is defective”.
1. Overall probability of a defect
Using total probability:
\[ P(D) = 0.4 \times 0.02 + 0.35 \times 0.03 + 0.25 \times 0.05 \]
\[ P(D) = 0.008 + 0.0105 + 0.0125 = 0.031 \]
So the overall probability of a defective part is 0.031, or 3.1%.
2. Probability that a defective part came from \(M_3\)
By Bayes’ theorem:
\[ P(M_3 \mid D) = \frac{P(D \mid M_3)P(M_3)}{P(D)} = \frac{0.25 \times 0.05}{0.031} \]
\[ P(M_3 \mid D) = \frac{0.0125}{0.031} \approx 0.403 \]
So the probability is about 0.403, or 40.3%.