Total probability theorem

A defective item could have come from any of three machines. A drawn ball could have come from either of two bags. In each case the outcome we care about — "defective," "red" — can arrive by several different routes, and we know how likely each route is and how likely the outcome is given that route. The total probability theorem stitches these route-by-route probabilities into one overall probability, using nothing more than a partition of the sample space and the multiplication theorem already met earlier in the chapter. It is the single most useful computational tool in this part of CBSE Class 12 probability, and it is also the direct setup for Bayes' theorem, which asks the reverse question — given the outcome happened, which route was it likely to have come from?

The core insight

When an event can occur through several mutually exclusive, exhaustive routes $E_1,\dots,E_n$ , its overall probability is the weighted sum $P(A)=\sum_i P(E_i)P(A\mid E_i)$ , weighting each route's conditional chance of $A$ by how likely that route is.

The core idea

Partitions first. Everything here rests on $E_1, E_2, \dots, E_n$ being a genuine partition of the sample space $S$ : pairwise disjoint, together covering all of $S$ , and each with $P(E_i) > 0$ . Every outcome in $S$ belongs to exactly one $E_i$ — no overlap, no gaps.

The theorem. For any event $A$ of $S$ and a partition $E_1,\dots,E_n$ :

P(A) = \sum_{i=1}^{n} P(E_i)\,P(A\mid E_i).

where: $P(E_i)$ is the probability of route $i$ (a prior), $P(A\mid E_i)$ is the conditional probability that $A$ happens given route $i$ was taken, and the sum runs over every piece of the partition — no route is left out, and none is counted twice.

Why the partition condition matters. Because $E_1,\dots,E_n$ partition $S$ , the event $A$ splits cleanly as $A = (A\cap E_1)\cup\cdots\cup(A\cap E_n)$ , and these pieces are pairwise disjoint since the $E_i$ are. That disjointness is what lets us just add the pieces without correcting for overlap — if the $E_i$ were not a genuine partition, some outcomes would be double-counted or missed, and the formula would simply be wrong.

Figure

Two boxes labelled route one and route two feeding into a single outcome bucket with route probabilities and conditional probabilities marked on each arrow

The core idea, continued

The formula is really "break the outcome into routes, weight each route by how likely it is, weight the outcome by how likely it is on that route, add." This is why the theorem is sometimes called the law of total probability — it accounts for the outcome's total probability across every possible cause.

A tree diagram makes the bookkeeping automatic. Draw a first level of branches for $E_1, E_2, \dots, E_n$ , each labelled with $P(E_i)$ ; from every $E_i$ branch draw a second-level branch to $A$ , labelled with $P(A\mid E_i)$ . Multiplying along a path gives $P(E_i)P(A\mid E_i)$ , and summing every path that ends at $A$ gives exactly the total probability formula. This is the same picture students already use for conditional probability problems — the total probability theorem is simply "read the tree, multiply along each branch to $A$ , then add the branches."

Relation to what comes next. The total probability theorem answers "given the routes, what is the overall chance of $A$ ?" Bayes' theorem, covered immediately afterward in the same chapter, answers the reverse question — "given that $A$ has already happened, which route was it likely to have come from?" Both use the same partition and the same conditional probabilities $P(A\mid E_i)$ ; Bayes' theorem simply divides one term of the total-probability sum by the whole sum:

P(E_k\mid A) = \frac{P(E_k)P(A\mid E_k)}{\sum_{i=1}^n P(E_i)P(A\mid E_i)}.

Every quantity on the right is something the total probability theorem already computes, which is why examiners often ask both questions back to back on the same word problem — get the total probability calculation right and Bayes' theorem is almost free.

Derivation ATotal probability theorem for a partition

The goal is to prove, for any partition $E_1, E_2, \dots, E_n$ of a sample space $S$ (pairwise disjoint, exhaustive, each with $P(E_i) > 0$ ) and any event $A$ of $S$ , that $P(A) = \sum_{i=1}^n P(E_i)\,P(A\mid E_i)$ , using only the addition rule for disjoint events and the multiplication theorem of conditional probability already established earlier in the chapter.

Step 1. Split $A$ using the partition. Since $E_1,\dots,E_n$ cover all of $S$ with no overlap, every outcome of $A$ lies in exactly one $E_i$ . So $A = (A\cap E_1)\cup(A\cap E_2)\cup\cdots\cup(A\cap E_n)$ .

Step 2. Note the pieces are pairwise disjoint. Since $E_i \cap E_j = \varnothing$ for $i \ne j$ , it follows that $(A\cap E_i)\cap(A\cap E_j) = \varnothing$ too — the pieces $A\cap E_1, \dots, A\cap E_n$ share no outcomes.

Step 3. Apply the addition rule for disjoint events. Because the pieces from Step 2 are pairwise disjoint, $P(A) = P(A\cap E_1) + P(A\cap E_2) + \cdots + P(A\cap E_n) = \sum_{i=1}^n P(A\cap E_i)$ .

Step 4. Rewrite each term with the multiplication theorem. For each $i$ , the multiplication theorem gives $P(A\cap E_i) = P(E_i)\,P(A\mid E_i)$ . Substituting into Step 3 gives $P(A) = \sum_{i=1}^n P(E_i)\,P(A\mid E_i)$ , which is the total probability theorem.

Worked example

Bag I contains 4 red and 4 black balls, and Bag II contains 2 red and 6 black balls. One bag is selected at random (each bag equally likely), and one ball is drawn from it. Find the probability that the ball drawn is red.

Step 1. Set up the partition. Let $E_1$ = "Bag I is selected," $E_2$ = "Bag II is selected." Since one bag is chosen at random, $P(E_1) = P(E_2) = \dfrac12$ , and $\{E_1, E_2\}$ is a valid partition of "which bag is chosen."

Step 2. Find the conditional probabilities. Let $A$ = "the ball drawn is red." From Bag I (4 red out of 8): $P(A\mid E_1) = \dfrac{4}{8} = \dfrac12$ . From Bag II (2 red out of 8): $P(A\mid E_2) = \dfrac{2}{8} = \dfrac14$ .

Step 3. Apply the theorem. $P(A) = P(E_1)P(A\mid E_1) + P(E_2)P(A\mid E_2) = \dfrac12\times\dfrac12 + \dfrac12\times\dfrac14 = \dfrac14 + \dfrac18 = \dfrac38$ .

The probability the ball drawn is red is $\boxed{\dfrac38}$ .

Competitive stretch

Extend to three routes with the factory example: machines I, II, III supply 25%, 35%, 40% of output with defect rates 5%, 4%, 2%. Then $P(\text{defective}) = (0.25)(0.05)+(0.35)(0.04)+(0.40)(0.02) = 0.0125+0.0140+0.0080 = 0.0345$ , i.e. $3.45\%$ — same three-step structure, just one more term in the sum.

Common traps

Forgetting to check that $E_1,\dots,E_n$ genuinely form a partition before applying the theorem — overlapping or non-exhaustive events break the formula.
Mixing up $P(A\mid E_i)$ with $P(E_i\mid A)$ — total probability needs the "route to outcome" direction, not the reverse (that reverse direction is Bayes' theorem).
Dropping terms when there are three or more routes — every $E_i$ in the partition must contribute, however small its probability looks.
Mixing percentage and decimal forms mid-calculation — convert every $P(E_i)$ and $P(A\mid E_i)$ to the same format (usually decimals) before multiplying.

Ask the tutor: Stuck on identifying the right partition or conditional probabilities in a multi-bag or multi-machine word problem? Ask the AI tutor to walk through setting up the events before applying the formula.

How is the total probability theorem different from the multiplication theorem I just learned?

The multiplication theorem gives you a single term, $P(A\cap E_i) = P(E_i)P(A\mid E_i)$ — the probability of one specific route and outcome happening together. The total probability theorem adds up all such terms across every route in a partition to get the overall probability of $A$ regardless of which route it came through. You need the multiplication theorem as a building block inside the total probability theorem; they are not competing tools, one feeds the other.

Do I always need exactly two events in the partition, like two bags?

No, a partition can have any number of pieces $n \ge 2$ , as long as they are pairwise disjoint and exhaustive. The factory example with three machines works exactly the same way, just with three terms in the sum instead of two. Word problems with more sources, causes, or stages simply mean more terms to add, not a different method.

Why does the theorem fail if the events overlap or do not cover the whole sample space?

The derivation relies on splitting $A$ into disjoint pieces $A\cap E_1, \dots, A\cap E_n$ so that their probabilities can simply be added without double-counting. If the $E_i$ overlap, some outcomes of $A$ get counted in more than one term, inflating the sum. If they do not cover all of $S$ , some outcomes of $A$ are missed entirely, and the sum falls short. Always check pairwise-disjoint-and-exhaustive before trusting the formula.