The image provides a formula for calculating the probability of an event $A$, denoted as $P(A)$, using the law of total probability. The formula expresses $P(A)$ as the sum of the products of conditional probabilities $P(A|B_i)$ and probabilities $P(B_i)$ for a set of mutually exclusive and exhaustive events $B_1, B_2, ..., B_n$. The formula is: $P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + ... + P(A|B_n)P(B_n)$

1. Problem Description

The image provides a formula for calculating the probability of an event $A$, denoted as $P(A)$, using the law of total probability. The formula expresses $P(A)$ as the sum of the products of conditional probabilities $P(A|B_i)$ and probabilities $P(B_i)$ for a set of mutually exclusive and exhaustive events $B_1, B_2, ..., B_n$. The formula is:

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

2. Solution Steps

The provided formula is a statement of the law of total probability. It states that if $B_1, B_2, ..., B_n$ are mutually exclusive and exhaustive events, then the probability of event $A$ can be calculated by summing the products of the conditional probability of $A$ given each $B_i$ and the probability of each $B_i$. Mathematically, this can be written as:

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

This is because the events $A \cap B_1, A \cap B_2, ..., A \cap B_n$ are mutually exclusive and their union is equal to A.

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

Using the conditional probability definition, $P(A \cap B_i) = P(A|B_i)P(B_i)$.

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

3. Final Answer

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