The Law of Total Probability

Intuition first

Sometimes a probability is awkward to compute directly but easy to compute once you know which situation you are in.

"What is the chance a randomly chosen visitor buys something?" is hard to answer in one step. But split visitors into mobile and desktop, and each piece becomes easy: mobile visitors buy 20% of the time, desktop visitors 40% of the time. If 60% of traffic is mobile, the overall rate is the weighted average

$$0.20 \times 0.60 \;+\; 0.40 \times 0.40 \;=\; 0.28$$

That is the whole law. Break the world into non-overlapping cases, solve each, then recombine weighted by how likely each case is.

The only thing that ever goes wrong is the splitting: the cases must not overlap, and together they must cover everything.

Partitions

A collection of events $B_1, B_2, \dots, B_n$ is a partition of the sample space when two conditions hold:

$$B_i \cap B_j = \varnothing \quad \text{for all } i \neq j \qquad \text{(no overlap)}$$ $$B_1 \cup B_2 \cup \dots \cup B_n = \Omega \qquad \text{(covers everything)}$$

Exactly one of the $B_i$ occurs, always. The simplest partition is any event and its complement: $\{B, B^c\}$.

The law

For any event $A$ and any partition $B_1, \dots, B_n$ with $\Prob(B_i) > 0$:

$$\Prob(A) = \sum_{i=1}^{n} \Prob(A \mid B_i)\,\Prob(B_i)$$

The two-case version, used constantly:

$$\Prob(A) = \Prob(A \mid B)\Prob(B) + \Prob(A \mid B^c)\Prob(B^c)$$

Derivation from the axiomsAdvanced

Because the $B_i$ partition $\Omega$, the event $A$ can be written as $A$ intersected with the whole space, then distributed:

$$A = A \cap \Omega = A \cap (B_1 \cup B_2 \cup \dots \cup B_n)$$$$= (A \cap B_1) \cup (A \cap B_2) \cup \dots \cup (A \cap B_n)$$

These pieces are pairwise disjoint. If $i \neq j$ then

$$(A \cap B_i) \cap (A \cap B_j) = A \cap (B_i \cap B_j) = A \cap \varnothing = \varnothing$$

so Axiom 3 (additivity) applies:

$$\Prob(A) = \sum_{i=1}^{n} \Prob(A \cap B_i)$$

Finally rewrite each joint using the multiplication rule, $\Prob(A \cap B_i) = \Prob(A \mid B_i)\Prob(B_i)$:

$$\Prob(A) = \sum_{i=1}^{n} \Prob(A \mid B_i)\,\Prob(B_i)$$

Two ingredients only: disjointness lets you add, and the multiplication rule converts joints into conditionals.

A useful sanity check

Because the weights $\Prob(B_i)$ sum to 1, the result is a genuine weighted average. It must therefore lie between the smallest and largest of the conditional probabilities:

$$\min_i \Prob(A \mid B_i) \;\leq\; \Prob(A) \;\leq\; \max_i \Prob(A \mid B_i)$$

In the shop example, $0.28$ sits between $0.20$ and $0.40$. If your answer falls outside that range, the arithmetic is wrong — this catches errors instantly.

Solved problem 1 · Three factories

A company sources components from three factories. Factory A supplies 50% of components with a 2% defect rate; factory B supplies 30% with a 3% defect rate; factory C supplies 20% with a 5% defect rate. A component is picked at random. What is the probability it is defective?

Step 1 — identify the partition and check it

The factories partition the components: a component comes from exactly one factory, and the three cover all supply.

$$\Prob(A) = 0.50, \qquad \Prob(B) = 0.30, \qquad \Prob(C) = 0.20$$

Check exhaustiveness: $0.50 + 0.30 + 0.20 = 1.00 \;\checkmark$

Step 2 — write the conditional defect rates

Let $D$ = "the component is defective".

$$\Prob(D \mid A) = 0.02, \qquad \Prob(D \mid B) = 0.03, \qquad \Prob(D \mid C) = 0.05$$

Step 3 — apply the law term by term

$$\Prob(D) = \Prob(D \mid A)\Prob(A) + \Prob(D \mid B)\Prob(B) + \Prob(D \mid C)\Prob(C)$$$$= 0.02(0.50) + 0.03(0.30) + 0.05(0.20)$$$$= 0.010 + 0.009 + 0.010$$$$= 0.029$$

Step 4 — sanity check

The conditional rates range from 0.02 to 0.05, and $0.029$ lies inside that range $\checkmark$

It sits near the low end, as expected: the largest supplier also has the lowest defect rate.

Answer

$\Prob(D) = 0.029$, that is 2.9%.

Solved problem 2 · A two-stage experiment

Urn I contains 4 red and 6 blue balls. Urn II contains 7 red and 3 blue balls. A fair coin is tossed: on heads we draw from urn I, on tails from urn II. What is the probability the drawn ball is red?

Step 1 — the partition is the coin toss

$$\Prob(\text{urn I}) = \Prob(\text{urn II}) = 0.5$$

These are mutually exclusive and exhaustive, so they form a valid partition.

Step 2 — conditional probabilities within each urn

$$\Prob(R \mid \text{I}) = \frac{4}{10} = 0.4, \qquad \Prob(R \mid \text{II}) = \frac{7}{10} = 0.7$$

Step 3 — combine

$$\Prob(R) = 0.4(0.5) + 0.7(0.5) = 0.20 + 0.35 = 0.55$$

Step 4 — check

With equal weights the answer must be the plain average of 0.4 and 0.7, which is $0.55 \;\checkmark$

Answer

$\Prob(\text{red}) = 0.55$.

Why this matters

The law of total probability is not a standalone trick. It is the denominator of Bayes' theorem, and it appears throughout the curriculum:

• Bayes' theorem uses it to compute $\Prob(A)$, the normalising constant.
• Marginalisation in probabilistic models is this law: summing a joint distribution over the values of a hidden variable.
• Variable elimination in Bayesian networks applies it repeatedly, which is exactly why that algorithm is tractable.
• Mixture models define $\Prob(x) = \sum_k \Prob(x \mid k)\Prob(k)$ — a Gaussian mixture is the law of total probability with Gaussian components.

Exercise 1

A student passes an exam with probability 0.9 if they studied and 0.3 if they did not. 70% of students study. What fraction of students pass?

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Partition on studying, $S$ and $S^c$, with $\Prob(S) = 0.7$ and $\Prob(S^c) = 0.3$ (these sum to 1 ✓).

$$\Prob(\text{pass}) = 0.9(0.7) + 0.3(0.3) = 0.63 + 0.09 = 0.72$$

Check: $0.72$ lies between $0.3$ and $0.9$ ✓, and closer to $0.9$ because most students study.

Exercise 2

A box holds 3 fair coins and 1 biased coin that lands heads with probability 0.8. A coin is chosen at random and tossed. Find the probability of heads.

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Partition on which coin was chosen:

$$\Prob(\text{fair}) = \tfrac34, \qquad \Prob(\text{biased}) = \tfrac14$$$$\Prob(H) = 0.5\left(\tfrac34\right) + 0.8\left(\tfrac14\right) = 0.375 + 0.200 = 0.575$$

Check: between 0.5 and 0.8 ✓, near the low end since fair coins dominate.

Exercise 3

A student writes: "60% of emails are from known senders and 25% of emails are spam, so $\Prob(\text{spam}) = \Prob(\text{spam} \mid \text{known})(0.6) + \Prob(\text{spam} \mid \text{unknown})(0.25)$." Identify the error.

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The weights must be the probabilities of the partition cases, not a mix of a case probability and the very quantity being computed.

The partition is $\{\text{known}, \text{unknown}\}$, so the weights are $0.6$ and $1 - 0.6 = 0.4$:

$$\Prob(\text{spam}) = \Prob(\text{spam} \mid \text{known})(0.6) + \Prob(\text{spam} \mid \text{unknown})(0.4)$$

The $0.25$ is $\Prob(\text{spam})$ itself — the left-hand side. Using it as a weight is circular, and since $0.6 + 0.25 = 0.85 \neq 1$ the weights do not even form a partition, which is the quickest way to spot the mistake.

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Next: Bayes' Theorem — which is this law run backwards, and the single most useful result in the module.