Learning Objectives

By the end of this chapter, you will be able to:

  • Calculate conditional probabilities
  • Apply the multiplication rule for independent events
  • Apply the multiplication rule for dependent events
  • Distinguish between independent and dependent events
  • Solve problems involving “and” probabilities

Multiplication Rule Overview

The multiplication rule is used to find the probability of event A AND event B occurring together.

This answers questions like:

  • What’s the probability of drawing two Kings in a row?
  • What’s the probability of passing both exams?
  • What’s the probability of selecting two defective items?

Conditional Probability

Definition

Conditional probability is the probability of an event occurring, given that another event has already occurred.

\[P(A|B) = \text{"Probability of A given B"}\]

Formula

\[P(A|B) = \frac{P(A \cap B)}{P(B)}\]

Where:

  • $P(A B)$ = Probability of A given B has occurred
  • $P(A \cap B)$ = Probability of both A and B
  • $P(B)$ = Probability of B (must be > 0)
flowchart LR
    A["Total Sample Space"]
    B["Event B has occurred"]
    C["Now what's P(A)?"]
    D["Only consider outcomes where B is true"]

    A --> B --> C --> D

Example 1: Simple Conditional Probability

Problem: A card is drawn and shown to be red. What is the probability it’s a Heart?

Solution:

  • Event A = Heart
  • Event B = Red card (given)

Since we know the card is red (26 cards), and 13 of those are hearts:

\[P(\text{Heart}|\text{Red}) = \frac{13}{26} = \frac{1}{2} = 0.5\]

Step-by-Step Example 2: Survey Data

Problem: A survey of 200 employees shows:

  Male Female Total
Satisfied 70 50 120
Dissatisfied 30 50 80
Total 100 100 200

Find: (a) P(Satisfied | Male) (b) P(Male | Satisfied) (c) P(Female | Dissatisfied)

Solution:

**(a) P(Satisfied Male)**

Given the employee is male, what’s the probability of being satisfied?

\[P(\text{Satisfied}|\text{Male}) = \frac{\text{Male and Satisfied}}{\text{Total Males}} = \frac{70}{100} = 0.70\]
**(b) P(Male Satisfied)**

Given the employee is satisfied, what’s the probability of being male?

\[P(\text{Male}|\text{Satisfied}) = \frac{\text{Male and Satisfied}}{\text{Total Satisfied}} = \frac{70}{120} = 0.583\]
**(c) P(Female Dissatisfied)**
\[P(\text{Female}|\text{Dissatisfied}) = \frac{\text{Female and Dissatisfied}}{\text{Total Dissatisfied}} = \frac{50}{80} = 0.625\]

Independent vs Dependent Events

Independent Events

Events A and B are independent if the occurrence of one does NOT affect the probability of the other.

$$P(A B) = P(A)$$
$$P(B A) = P(B)$$

Examples of Independent Events:

  • Rolling a die twice
  • Flipping a coin multiple times
  • Selecting items WITH replacement

Dependent Events

Events A and B are dependent if the occurrence of one AFFECTS the probability of the other.

\[P(A|B) \neq P(A)\]

Examples of Dependent Events:

  • Drawing cards WITHOUT replacement
  • Selecting items from a batch without replacement
  • Sequential selections from a finite population
flowchart TD
    A[Are events independent?]
    A -->|Yes| B["P(A and B) = P(A) × P(B)"]
    A -->|No| C["P(A and B) = P(A) × P(B|A)"]

    D["With replacement ➔ Independent"]
    E["Without replacement ➔ Dependent"]

Multiplication Rule: Two Cases

Case 1: Independent Events

\[P(A \cap B) = P(A) \times P(B)\]

Case 2: Dependent Events

\[P(A \cap B) = P(A) \times P(B|A)\]

Or equivalently:

\[P(A \cap B) = P(B) \times P(A|B)\]

Step-by-Step Example 3: Independent Events (Coin Toss)

Problem: A fair coin is tossed twice. Find the probability of getting two heads.

Solution:

Events are independent (first toss doesn’t affect second):

\(P(\text{HH}) = P(\text{H on 1st}) \times P(\text{H on 2nd})\) \(= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25\)

Step-by-Step Example 4: Independent Events (Die Roll)

Problem: A die is rolled three times. Find the probability of getting: (a) Three 6s (b) No 6s

Solution:

(a) Three 6s:

\(P(\text{6, 6, 6}) = P(6) \times P(6) \times P(6)\) \(= \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} = 0.0046\)

(b) No 6s:

\(P(\text{No 6}) = P(\text{Not 6}) \times P(\text{Not 6}) \times P(\text{Not 6})\) \(= \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216} = 0.579\)

Step-by-Step Example 5: Dependent Events (Cards)

Problem: Two cards are drawn WITHOUT replacement. Find the probability of getting two Kings.

Solution:

Step 1: Probability of first King \(P(\text{1st King}) = \frac{4}{52}\)

Step 2: Probability of second King GIVEN first was King

After drawing one King: 3 Kings left, 51 cards left \(P(\text{2nd King}|\text{1st King}) = \frac{3}{51}\)

Step 3: Apply multiplication rule for dependent events \(P(\text{Two Kings}) = P(\text{1st King}) \times P(\text{2nd King}|\text{1st King})\) \(= \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \approx 0.0045\)

Step-by-Step Example 6: With vs Without Replacement

Problem: A box contains 6 red and 4 blue balls. Two balls are drawn. Find the probability of getting two red balls: (a) With replacement (b) Without replacement

Solution:

(a) With Replacement (Independent):

\(P(\text{Red, Red}) = P(\text{Red}) \times P(\text{Red})\) \(= \frac{6}{10} \times \frac{6}{10} = \frac{36}{100} = 0.36\)

(b) Without Replacement (Dependent):

\(P(\text{Red, Red}) = P(\text{1st Red}) \times P(\text{2nd Red}|\text{1st Red})\) \(= \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \approx 0.333\)

Step-by-Step Example 7: Quality Control

Problem: A batch of 20 products contains 4 defective items. If 2 items are randomly selected without replacement, find the probability that: (a) Both are defective (b) Both are non-defective (c) At least one is defective

Solution:

(a) Both defective:

\[P(\text{D, D}) = \frac{4}{20} \times \frac{3}{19} = \frac{12}{380} = \frac{3}{95} = 0.0316\]

(b) Both non-defective:

Non-defective = 20 - 4 = 16

\[P(\text{ND, ND}) = \frac{16}{20} \times \frac{15}{19} = \frac{240}{380} = \frac{12}{19} = 0.632\]

(c) At least one defective:

\(P(\text{At least 1 D}) = 1 - P(\text{Both ND})\) \(= 1 - 0.632 = 0.368\)

General Multiplication Rule for Multiple Events

For three dependent events:

\[P(A \cap B \cap C) = P(A) \times P(B|A) \times P(C|A \cap B)\]

Example 8: Three Cards

Problem: Three cards are drawn without replacement. Find the probability of getting three Aces.

Solution:

\[P(\text{3 Aces}) = P(\text{1st Ace}) \times P(\text{2nd Ace}|\text{1st}) \times P(\text{3rd Ace}|\text{1st two})\] \[= \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50}\] \[= \frac{24}{132600} = \frac{1}{5525} \approx 0.000181\]

Testing for Independence

Method: Compare P(A|B) with P(A)

If $P(A B) = P(A)$, events are independent.

Alternative: Check if P(A∩B) = P(A)×P(B)

Example 9: Testing Independence

Problem: From the employee survey:

  Male Female Total
Manager 20 10 30
Staff 80 90 170
Total 100 100 200

Are gender and position independent?

Solution:

Step 1: Calculate individual probabilities \(P(\text{Manager}) = \frac{30}{200} = 0.15\) \(P(\text{Male}) = \frac{100}{200} = 0.50\)

Step 2: Calculate conditional probability \(P(\text{Manager}|\text{Male}) = \frac{20}{100} = 0.20\)

Step 3: Compare \(P(\text{Manager}|\text{Male}) = 0.20 \neq 0.15 = P(\text{Manager})\)

Conclusion: Gender and position are NOT independent (they are dependent). Males are more likely to be managers.

Combined Application: Addition and Multiplication Rules

Example 10: Complex Problem

Problem: A company has two branches. Branch A has 60% success rate on projects, Branch B has 70% success rate. A project is equally likely to be assigned to either branch. What is the probability of project success?

Solution:

flowchart TD
    A[Project Assigned] --> B["Branch A<br/>P = 0.5"]
    A --> C["Branch B<br/>P = 0.5"]
    B --> D["Success<br/>P = 0.6"]
    B --> E["Failure<br/>P = 0.4"]
    C --> F["Success<br/>P = 0.7"]
    C --> G["Failure<br/>P = 0.3"]

Using Total Probability:

\[P(\text{Success}) = P(\text{A}) \times P(\text{Success}|\text{A}) + P(\text{B}) \times P(\text{Success}|\text{B})\] \[= 0.5 \times 0.6 + 0.5 \times 0.7\] \[= 0.30 + 0.35 = 0.65\]

Answer: 65% probability of success

Summary Table: When to Use Which Rule

Situation Rule Formula
A OR B (mutually exclusive) Addition $P(A) + P(B)$
A OR B (not mutually exclusive) General Addition $P(A) + P(B) - P(A \cap B)$
A AND B (independent) Simple Multiplication $P(A) \times P(B)$
A AND B (dependent) General Multiplication $P(A) \times P(B \mid A)$

Practice Problems

Problem 1

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find: (a) P(both red) (b) P(both blue) (c) P(one of each color)

Problem 2

The probability of rain on any day in monsoon is 0.4. Assuming independence, find the probability of rain on: (a) Both Monday and Tuesday (b) At least one of Monday or Tuesday (c) Neither day

Problem 3

From the table below, test whether department and gender are independent:

  Male Female Total
Admin 30 20 50
Technical 40 10 50
Total 70 30 100

Problem 4

Two candidates apply for a job. P(A gets interview) = 0.7, P(B gets interview) = 0.6. If these events are independent, find: (a) P(both get interviews) (b) P(neither gets interview) (c) P(at least one gets interview)

Problem 5

A fair die is rolled until a 6 appears. What is the probability that it takes exactly 3 rolls?

Summary

Concept Formula Key Point
Conditional Probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$ Probability given prior info
Independence Test $P(A|B) = P(A)?$ Events don’t affect each other
Multiplication (Ind.) $P(A \cap B) = P(A) \times P(B)$ With replacement
Multiplication (Dep.) $P(A \cap B) = P(A) \times P(B|A)$ Without replacement

Next Topic

In the next chapter, we will study Probability Distributions - starting with the Binomial Distribution, which models the number of successes in repeated independent trials.