When we talk about total probability we are talking about a partition space. A partition of any sample space is nothing but simply a collection of events that are disjoint say $C_1, C_2…, C_k$ and the union of all these events is equal to the sample space.

Such kind of partition will also divide any set say B of the sample space into disjoint pieces.

Let us suppose that we have an event B and $A_1, A_2,…, A_k$ are the partitions then, event B is the union of these partitions such that:

$B$ = $(B \cap A_1) \cup (B \cap A_2) \cup … \cup (B \cap A_k)$

Since all these partitions are disjoint so we have

$P(B)$ = $P (B \cap A_1) + P (B \cap A_2) + … + P (B \cap A_k)$

We know by the multiplication rule that:

$P(B \cap A_1)$ = $P (A_1) P (B|A_1)$

where $P (B|A_1)$ is the conditional probability which gives the probability of occurrence of event B when event $A_1$ has already occurred.

Applying this rule above we get,

$P (B)$ = $P (A_1) P (B|A_1) + P (A_2) P (B|A_2) + … + P (A_k) P (B|A_k)$

This is the law of total probability. This law is very useful in many places. For example: We are drawing two cards from a deck of cards. The first card drawn is a king. To find the probability if the second one is an ace or not we use the law of total probability only which makes it easier to find.

**The law of total probability is also referred to as total probability theorem or law of alternatives.**

Let us check some examples of problem solving that involves the total probability theorem.

## Example

**Let us check some examples of problem solving that involves the total probability theorem.**

**Example 1:** We draw two cards from a deck of shuffled cards without replacement. Find the probability of getting the second card a queen.

**Solution:** Let $‘A’$ represent the event of getting the first card a queen.

Let $‘B’$ represent the event that the first card is not a queen.

Let $‘E’$ represent the event that the second card is a queen.

Then the probability that the second card will be a queen or not will be represented by the law of total probability as:

$P (E)$ = $P (A) P (E|A) + P (B) P (E|B)$

Where $P (E)$ is the probability that second card is a queen

$P (A)$ is the probability that the first card is a queen

$P (E|A)$ is the probability that the second card is a queen given that first card is a queen.

$P (B)$ is the probability that the first card is not a queen

$P (E|B)$ is the probability that the second card is a queen but the first card drawn is not a queen.

According to question:

$P (A)$ = $\frac{4}{52}$ $P (B)$ = $\frac{48}{52}$ $P (E|A)$ = $\frac{3}{51}$ $P (E|B)$ = $\frac{4}{51}$

On substituting we get,

$P (E)$ = $\frac{4}{52}$ * $\frac{3}{51}$ + $\frac{48}{52}$ * $\frac{4}{51}$

= $\frac{(3 * 4)}{(52 * 51)}$ + $\frac{(48 * 4)}{(52 * 51)}$

= $\frac{(4 * (3 + 48))}{(52 * 51)}$

= $\frac{(4 * 51)}{ (52 * 51)}$

= $\frac{4}{52}$

= $\frac{1}{13}$