Probability theory is essential in banking and finance as it provides a mathematical framework to model uncertainty and risk. Financial institutions rely on probability to assess credit risk, market fluctuations, and investment outcomes. This chapter introduces fundamental concepts of probability events and their classifications, equipping banking professionals with analytical tools to make informed decisions under uncertainty.

7.1 Probability events

Probability events represent outcomes or sets of outcomes from a random process, such as loan approvals or market movements. Understanding how to define and calculate probabilities of events is critical for risk assessment and decision-making in finance.

7.1.1 Definition of Probability Event

A probability event is a specific outcome or a collection of outcomes from a sample space of a random experiment. The sample space, denoted by \( S \), contains all possible outcomes, and an event \( E \) is any subset of \( S \).

The probability of an event \( E \), represented as \( P(E) \), is a measure between 0 and 1 that quantifies the likelihood of \( E \) occurring.

The basic properties include:

• \( 0 \leq P(E) \leq 1 \) for any event \( E \)
• \( P(S) = 1 \) since some outcome in the sample space must occur
• \( P(\emptyset) = 0 \) where \( \emptyset \) is the empty event

$$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in } S}$$

Worked Examples

Example 1: A bank issues 100 loan applications, 40 are approved. Find the probability that a randomly selected loan is approved.

Given: Number of approved loans = 40, Total loans = 100

$$P(\text{approved}) = \frac{40}{100}$$

$$= 0.4$$

Answer: 0.4 (40%)

Example 2: A SACCO has 5 types of shares. If one share type is randomly selected, find the probability it is a preferred share, given there are 2 preferred share types.

Given: Preferred share types = 2, Total share types = 5

$$P(\text{preferred}) = \frac{2}{5}$$

$$= 0.4$$

Answer: 0.4 (40%)

Example 3: In a survey of 200 customers, 50 have savings accounts. Find the probability that a customer selected at random does not have a savings account.

Given: Customers with savings = 50, Total customers = 200

$$P(\text{no savings}) = 1 - P(\text{savings}) = 1 - \frac{50}{200}$$

$$= 1 - 0.25 = 0.75$$

Answer: 0.75 (75%)

Example 4: A Kenyan retail bank randomly selects 3 customers from a group of 10, where 7 have active loans. Find the probability the first selected customer has an active loan.

Given: Active loan customers = 7, Total customers = 10

$$P(\text{active loan}) = \frac{7}{10}$$

$$= 0.7$$

Answer: 0.7 (70%)

Example 5: A financial institution offers 4 types of insurance policies. If a policyholder is randomly selected, find the probability they hold either life or health insurance, given 1 is life and 1 is health.

Given: Life policies = 1, Health policies = 1, Total policies = 4

$$P(\text{life or health}) = \frac{1 + 1}{4} = \frac{2}{4}$$

$$= 0.5$$

Answer: 0.5 (50%)

7.1.2 Sample Space and Event Sets

The sample space \( S \) is the set of all possible outcomes of an experiment, such as all possible credit ratings assigned to loan applicants. Events are subsets of \( S \).

The union, intersection, and complement of events are important operations:

• Union \( A \cup B \): Event that either \( A \) or \( B \) or both occur.
• Intersection \( A \cap B \): Event that both \( A \) and \( B \) occur.
• Complement \( A^c \): Event that \( A \) does not occur.

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

$$P(A^c) = 1 - P(A)$$

Worked Examples

Example 1: In a bank, 60% of customers have savings accounts, 40% have loans, and 25% have both. Find the probability a customer has savings or loans.

Given: \( P(S) = 0.6 \), \( P(L) = 0.4 \), \( P(S \cap L) = 0.25 \)

$$P(S \cup L) = 0.6 + 0.4 - 0.25$$

$$= 1.0 - 0.25 = 0.75$$

Answer: 0.75 (75%)

Example 2: A SACCO has 30% members with fixed deposits. Find the probability a member does not have a fixed deposit.

Given: \( P(FD) = 0.3 \)

$$P(FD^c) = 1 - 0.3 = 0.7$$

Answer: 0.7 (70%)

Example 3: At a bank, 50% of clients have mobile banking, 20% have internet banking, and 10% have both. Find the probability a client has mobile or internet banking.

Given: \( P(M) = 0.5 \), \( P(I) = 0.2 \), \( P(M \cap I) = 0.1 \)

$$P(M \cup I) = 0.5 + 0.2 - 0.1 = 0.6$$

Answer: 0.6 (60%)

Example 4: In a loan portfolio, 15% of loans are overdue. Find the probability a loan is not overdue.

Given: \( P(\text{overdue}) = 0.15 \)

$$P(\text{not overdue}) = 1 - 0.15 = 0.85$$

Answer: 0.85 (85%)

Example 5: A bank classifies customers into three risk categories: low (40%), medium (35%), and high (25%). Find the probability a customer is not medium risk.

Given: \( P(\text{medium}) = 0.35 \)

$$P(\text{not medium}) = 1 - 0.35 = 0.65$$

Answer: 0.65 (65%)

7.1.3 Calculating Probability of Events

Calculating probability involves counting favorable outcomes relative to total outcomes or using known probabilities and rules of addition, multiplication, and complements.

For equally likely outcomes:

$$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$

For non-equally likely outcomes, probabilities are assigned based on observed frequencies or subjective assessments.

Worked Examples

Example 1: A bank has 8 loan officers, 3 specialize in commercial loans. Find the probability a randomly selected officer specializes in commercial loans.

Given: Commercial loan officers = 3, Total officers = 8

$$P = \frac{3}{8} = 0.375$$

Answer: 0.375 (37.5%)

Example 2: A SACCO has 50 members, 20 have savings accounts, 10 have fixed deposits. Find the probability a member has savings or fixed deposit, assuming 5 have both.

Given: \( P(S) = \frac{20}{50} = 0.4 \), \( P(FD) = \frac{10}{50} = 0.2 \), \( P(S \cap FD) = \frac{5}{50} = 0.1 \)

$$P(S \cup FD) = 0.4 + 0.2 - 0.1 = 0.5$$

Answer: 0.5 (50%)

Example 3: A bank randomly selects a day in a 30-day month. The probability it rains on any day is 0.2. Find the probability it does not rain on the selected day.

Given: \( P(\text{rain}) = 0.2 \)

$$P(\text{no rain}) = 1 - 0.2 = 0.8$$

Answer: 0.8 (80%)

Example 4: In a credit portfolio, 70% loans are performing, 15% are non-performing, and 15% are under watch. Find the probability a loan is performing or under watch.

Given: \( P(\text{performing}) = 0.7 \), \( P(\text{under watch}) = 0.15 \)

Assuming no overlap,

$$P = 0.7 + 0.15 = 0.85$$

Answer: 0.85 (85%)

Example 5: A bank has 12 branches, 4 are in Nairobi. Find the probability a randomly chosen branch is outside Nairobi.

Given: Nairobi branches = 4, Total branches = 12

$$P(\text{outside Nairobi}) = 1 - \frac{4}{12} = 1 - \frac{1}{3} = \frac{2}{3}$$

$$= 0.6667$$

Answer: 0.6667 (66.67%)

Chapter Summary

This chapter introduced the concept of probability events as the foundation for measuring the likelihood of outcomes in various scenarios. It explored different types of events including simple events with single outcomes, compound events involving multiple outcomes, mutually exclusive events that cannot occur simultaneously, independent events where one event does not affect the other, and dependent events where outcomes influence each other. The chapter then detailed the fundamental rules of probability that govern how probabilities are calculated and combined. Bayes' Theorem was presented as a powerful tool for updating probabilities based on new information. The method of drawing probability trees was explained to visually represent and analyze sequences of events. Finally, the chapter demonstrated practical applications of probability theory in solving real-world problems, reinforcing the importance of these concepts in business mathematics and statistics.