Law of total probability: partition to count

The idea

To compute $P(A)$, split the sample space into pieces $B_1, B_2, \ldots, B_n$ forming a partition:

• Pairwise incompatible: $B_i \cap B_j = \emptyset$
• Union equal to $\Omega$: $B_1 \cup B_2 \cup \ldots \cup B_n = \Omega$

Then:

$P(A) = P(A \cap B_1) + P(A \cap B_2) + \ldots + P(A \cap B_n)$

Conditional version:

$P(A) = P(A | B_1) P(B_1) + P(A | B_2) P(B_2) + \ldots + P(A | B_n) P(B_n)$

Case $n = 2$: partition $B, \overline{B}$

$P(A) = P(A \cap B) + P(A \cap \overline{B}) = P(A|B) P(B) + P(A|\overline{B}) P(\overline{B})$

The most common version in high school.

Example: screening test

A disease affects $2 \%$ of the population. A test:

• Positive if sick: $99 \%$
• Positive if healthy: $5 \%$ (false positives)

Question: what’s the probability a random test is positive?

Let $M$ = “sick”, $T$ = “test positive”.

$P(M) = 0.02$, $P(\overline{M}) = 0.98$ $P(T | M) = 0.99$, $P(T | \overline{M}) = 0.05$

Law of total probability:

$P(T) = P(T|M) P(M) + P(T|\overline{M}) P(\overline{M}) = 0.99 \times 0.02 + 0.05 \times 0.98$

$P(T) = 0.0198 + 0.049 = 0.0688 \approx 6.9 \%$

On a tree diagram

Each path contributes with the product of its branches. Summing all paths leading to $A$ gives $P(A)$.

That’s exactly the law of total probability visualized.

Common mistakes

1. Adding without conditioning: $P(A) \neq P(A | B) + P(A | \overline{B})$.
2. Forgetting to multiply by $P(B_i)$ in the conditional version.
3. Using a “partition” that isn’t (union $\neq \Omega$).
4. Confusing $P(A \cap B)$ and $P(A | B)$.

?À toi de jouer

P(A|B) = 0.3, P(A|B̄) = 0.6, P(B) = 0.4. P(A) = ?

A.0.48B.0.90C.0.45D.0.18

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