Probability

Probability

Syllabus tag: KASNEB CPA | Foundation Level | CA15 Quantitative Analysis | Topic 6 Probability

Lesson objectives

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

• Apply the addition and multiplication rules
• Distinguish mutually exclusive from independent events
• Compute conditional probability
• Apply Bayes' theorem
• Compute an expected value and use it in a decision

Why this matters

Probability is the language of risk, and every later topic that deals with uncertainty — distributions, sampling, decision theory — is built on the four rules below.

The basics

P(A) = number of favourable outcomes / total possible outcomes

Probability lies between 0 (impossible) and 1 (certain).

The complement: P(not A) = 1 − P(A). This is often the quickest route — "at least one" problems are usually solved as 1 minus the probability of none.

The addition rule

Mutually exclusive events cannot both occur:

P(A or B) = P(A) + P(B)

With P(A) = 0.35 and P(B) = 0.28: 0.63

Not mutually exclusive events can overlap, so the overlap must be removed or it is counted twice:

P(A or B) = P(A) + P(B) − P(A and B)

With P(A) = 0.40, P(B) = 0.25 and P(A and B) = 0.10: 0.40 + 0.25 − 0.10 = 0.55

The multiplication rule

Independent events — one does not affect the other:

P(A and B) = P(A) × P(B) = 0.40 × 0.25 = 0.10

Dependent events:

P(A and B) = P(A) × P(B | A)

The distinction candidates confuse. Mutually exclusive means the events cannot happen together. Independent means one happening does not change the probability of the other. They are not the same, and in fact mutually exclusive events are never independent — if A occurs, the probability of B becomes zero, which is very much an effect.

Conditional probability

P(A | B) = P(A and B) / P(B)

= 0.10 / 0.25 = 0.40

Read it as: given that B has happened, the probability of A is 0.40.

Note that P(A) was 0.40 as well, so B tells us nothing about A here. When P(A | B) = P(A), the events are independent — that equality is the formal test.

Bayes' theorem

Used to revise a probability once new evidence arrives:

P(A | B) = P(B | A) × P(A) / P(B)

A machine produces 3% defective items. A test detects 95% of defects but also flags 4% of good items. If an item is flagged, what is the probability it is truly defective?

	Working
P(flagged and defective)	0.03 × 0.95	0.0285
P(flagged and good)	0.97 × 0.04	0.0388
P(flagged)		0.0673

P(defective | flagged) = 0.0285 / 0.0673 = 0.4235

The result is worth pausing on. Even with a test that catches 95% of defects, a flagged item is more likely to be good than defective — because good items are so much more numerous that 4% of them outweighs 95% of a small defective population. This is why screening tests for rare conditions produce so many false positives, and it is a standard examination point.

:::checkpoint The defect rate rises from 3% to 20% with the same test. Recompute the probability that a flagged item is truly defective, and explain in one sentence why the answer changes so much. :::

Expected value

EV = Σ (probability × outcome)

A project returns KES 40,000 with probability 0.3, KES 25,000 with probability 0.5, and a loss of KES 15,000 with probability 0.2:

EV = (0.3 × 40,000) + (0.5 × 25,000) + (0.2 × −15,000) = 12,000 + 12,500 − 3,000 = KES 21,500

Two cautions apply, and both are examinable:

• The expected value is often not a possible outcome. Here no scenario produces 21,500.
• It says nothing about the spread. A 20% chance of a loss is invisible in the single figure, and for a one-off decision that could matter more than the average.

Expected values suit repeated decisions, where outcomes average out over many trials.

Counting

Permutations — order matters: ⁿPᵣ = n! / (n − r)! Combinations — order does not: ⁿCᵣ = n! / [r!(n − r)!]

A committee is a combination; a ranked shortlist is a permutation. Choosing 3 from 8: as a combination, 56; as a permutation, 336.

:::checkpoint An examiner asks for the probability that at least one of five machines fails, where each fails independently with probability 0.1. Explain why computing 1 minus the probability of no failures is easier than the direct route. :::