We need to prove the Law of Total Probability, which states that if events $B_1, B_2, ..., B_n$ form a partition of the sample space $S$, then the probability of any event $A$ can be expressed as: $P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + ... + P(A|B_n)P(B_n)$ or equivalently, $P(A) = \sum_{i=1}^{n} P(A|B_i)P(B_i)$. The image shows this formula, but with some notational errors. It should be $P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + ... + P(A|B_n)P(B_n)$.

1. Problem Description

We need to prove the Law of Total Probability, which states that if events $B_1, B_2, ..., B_n$ form a partition of the sample space $S$, then the probability of any event $A$ can be expressed as:

$P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + ... + P(A|B_n)P(B_n)$

or equivalently, $P(A) = \sum_{i=1}^{n} P(A|B_i)P(B_i)$.

The image shows this formula, but with some notational errors. It should be $P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + ... + P(A|B_n)P(B_n)$.

2. Solution Steps

The events $B_1, B_2, ..., B_n$ form a partition of the sample space $S$. This means that they are mutually exclusive and their union is the entire sample space. Therefore,

$\bigcup_{i=1}^n B_i = S$

and $B_i \cap B_j = \emptyset$ for $i \ne j$.

We can write the event $A$ as the intersection of $A$ with the sample space $S$:

$A = A \cap S$

Since $\bigcup_{i=1}^n B_i = S$, we can substitute $S$ with $\bigcup_{i=1}^n B_i$:

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

Using the distributive property of intersection over union, we have:

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

Now, we want to find the probability of event $A$:

$P(A) = P(\bigcup_{i=1}^n (A \cap B_i))$

Since the $B_i$ are mutually exclusive, the events $A \cap B_i$ are also mutually exclusive. Therefore, the probability of the union is the sum of the probabilities:

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

We know the conditional probability formula:

$P(A|B_i) = \frac{P(A \cap B_i)}{P(B_i)}$

Rearranging this formula, we get:

$P(A \cap B_i) = P(A|B_i)P(B_i)$

Substituting this back into the equation for $P(A)$, we get:

$P(A) = \sum_{i=1}^n P(A|B_i)P(B_i)$

This can be written as:

$P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + ... + P(A|B_n)P(B_n)$

3. Final Answer

$P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + ... + P(A|B_n)P(B_n)$