48. If A is any event associated with sample space and if $E_1,E_2,E_3$ are mutually exclusive and exhaustive events, then which of the following are true?

(A) $P\left(A\right)=P\left(E_1\right)P\left(E_1\right|A)+P(E_2\left)P\right(E_2\left|A\right)+P\left(E_3\right)P\left(E_3\right|A)$

(B) $P\left(A\right)=P\left(A\right|E_1\left)P\right(E_1)+P(A\left|E_2\right)P\left(E_2\right)+P\left(A\right|E_3\left)P\right(E_3)$

(C) $P\left(E_i\right|A)=\frac{P}{\left(},i=1,2,3$

(D) $P\left(A\right|E_i)=\frac{P}{\left(},i=1,2,3$

$\mathrm{<br>}$

Choose the correct answer from the options given below:

The correct answer is (B) and (C) only.

To understand why statements (B) and (C) are correct, we can analyze the probability concepts of mutually exclusive and exhaustive events, the law of total probability, and Bayes' theorem.

1. Understanding the Events
We are given that $E_1$, $E_2$, and $E_3$ are mutually exclusive and exhaustive events.
- Mutually exclusive means that no two events can occur at the same time:
$E_i\cap E_j=\varnothing$ for $i\ne j$.
- Exhaustive means that their union forms the entire sample space $S$:
$E_1\cup E_2\cup E_3=S$.

For any event $A$ associated with the same sample space $S$, we can express $A$ as:
$A=A\cap S=A\cap (E_1\cup E_2\cup E_3)$
Using the distributive law of set theory, this becomes:
$A=(A\cap E_1)\cup (A\cap E_2)\cup (A\cap E_3)$

Since $E_1$, $E_2$, and $E_3$ are mutually exclusive, the intersections $(A\cap E_i)$ are also mutually exclusive. Therefore, the probability of event $A$ is the sum of their individual probabilities:
$P\left(A\right)=P(A\cap E_1)+P(A\cap E_2)+P(A\cap E_3)$

2. Evaluating Statement (B) - The Law of Total Probability
Using the multiplication rule of probability, the joint probability of two events can be written using conditional probability as:
$P(A\cap E_i)=P\left(A\right|E_i\left)P\right(E_i)$

Substituting this relation into the expression for $P\left(A\right)$, we obtain:
$P\left(A\right)=P\left(A\right|E_1\left)P\right(E_1)+P(A\left|E_2\right)P\left(E_2\right)+P\left(A\right|E_3\left)P\right(E_3)$
This is the Law of Total Probability and exactly matches the equation in (B). Thus, statement (B) is true.
(Note that statement (A) incorrectly defines $P\left(A\right)$ using the conditional terms $P\left(E_i\right|A)$ instead of $P\left(A\right|E_i)$, making statement (A) false.)

3. Evaluating Statement (C) - Bayes' Theorem
Bayes' Theorem allows us to find the conditional probability of an event $E_i$ given that event $A$ has occurred. By definition:
$P\left(E_i\right|A)=\frac{P(E_i\cap A)}{P\left(A\right)}=\frac{P\left(A\right|E_i\left)P\right(E_i)}{P\left(A\right)}$

Now, substituting the Law of Total Probability expression for $P\left(A\right)$ in the denominator, we get:
$P\left(E_i\right|A)=\frac{P\left(A\right|E_i\left)P\right(E_i)}{\sum _{j=1}^{3}P\left(A\right|E_j\left)P\right(E_j)},\text{for}i=1,2,3$
This matches statement (C). Thus, statement (C) is true.
(Note that statement (D) incorrectly sets up the ratio, making it false.)

Consequently, since only statements (B) and (C) are mathematically correct, the correct choice is (B) and (C) only.