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) $P(A) = P(E_1)P(E_1|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)$
(B) $P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)$
(C) $P(E_i|A)=\frac{P(A|E_i)P(E_i)}{Σ_{i=1}^3P(A|E_i)P(E_i)},i = 1,2,3$
(D) $P(A|E_i) =\frac{P(E_i|A)P(E_i)}{Σ_{i=1}^3P(E_i|A)P(E_i)},i = 1,2,3$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
Given $E_1,E_2,E_3$ are mutually exclusive and exhaustive
$E_1\cup E_2\cup E_3=S,\; P(E_1)+P(E_2)+P(E_3)=1$
Option (A)
$P(A)=P(E_1)P(E_1|A)+P(E_2)P(E_2|A)+P(E_3)P(E_3|A)$
This is incorrect since $P(E_i|A)$ appears instead of $P(A|E_i)$
Option (B)
$P(A)=P(A|E_1)P(E_1)+P(A|E_2)P(E_2)+P(A|E_3)P(E_3)$
This is the law of total probability
Correct
Option (C)
$P(E_i|A)=\frac{P(A|E_i)P(E_i)}{\sum_{j=1}^{3}P(A|E_j)P(E_j)},\; i=1,2,3$
This is Bayes’ theorem
Correct
Option (D)
$P(A|E_i)=\frac{P(E_i|A)P(E_i)}{\sum_{j=1}^{3}P(E_j|A)P(E_j)}$
This is not a valid probability identity
Incorrect
The correct statements are (B) and (C).