Reversing the condition, and counting

Bayes by table or tree — the branch over the total, no formula memorised — and ₙCₓ for the binomial ahead.

The question that runs the tree backwards

Georgian Bay Mutual's tree gave $P (fraud ∣ auto) = 0.10$ : the rate, given the type. The investigator's question is the reverse: a claim has been flagged fraudulent — what is the probability it was an auto claim? That is $P (auto ∣ fraud)$ , and the tree already holds every piece:

$P (auto ∣ fraud) = \frac{P ( auto \cap fraud )}{P ( fraud )} = \frac{0.060}{0.076} = 0.789$

The numerator is the one leaf where both events hold; the denominator is the total probability of the condition — the sum of every leaf ending in fraud (L5.5). Nothing new was needed: it is the definition of conditional probability with the tree supplying both parts.

Notice how different the two conditionals are. Ten percent of auto claims are flagged; but 79 % of flagged claims are auto — because auto claims are both the larger group and the higher-rate one. Swapping $P (A ∣ B)$ for $P (B ∣ A)$ is the most common conceptual error in this module.

The same thing by table

Give the tree a population of 1,000 claims: 600 auto (60 flagged, 540 not) and 400 property (16 flagged, 384 not). The flagged column totals 76, of which 60 are auto: $60/76 = 0.789$ . A table with invented counts is often the fastest route and the easiest to check.

The formula, for those who want it

Written in symbols, the reversal is Bayes' rule:

$P (B ∣ A) = \frac{P ( A ∣ B ) P ( B )}{P ( A )}$

with $P (A)$ expanded as the sum of branch products. IBS1 never states it and the paper never demands it — the tree in one line is all it says. Work the tree; write the formula only if you find it helps.

Counting the ways — combinations

Module 6's binomial asks, for example, "exactly two of twelve claims have an error". The two erroneous claims could be any two of the twelve, and the number of ways to choose them is a combination:

$(x n) =_{n} C_{x} = \frac{n !}{x ! ( n - x )!}$

where $n! = n \times (n - 1) \times \dots \times 1$ and $0! = 1$ . Order does not matter — choosing claims 3 and 7 is the same audit as choosing 7 and 3.

Maple Ledger's auditor picks three of ten invoices for a detailed check: $(3 10) = \frac{10 !}{3 ! 7 !} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ possible samples. Two of the fifty: $(2 50) = \frac{50 \times 49}{2} = 1, 225$ . Cancel the larger factorial before multiplying — $10! /7!$ is just $10 \times 9 \times 8$ — and the calculator never overflows. Python's math.comb and a calculator's nCr key give the same numbers.