A, B, C are mutually exclusive and exhaustive | vedclass

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(A) Given that $A, B, C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$. Given $P(A) = 0.30$ and $P(B) = 0.50$,we find

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$\frac{2}{35}$

Vedclass · Answer

(A) Given that $A, B, C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$. Given $P(A) = 0.30$ and $P(B) = 0.50$,we find $P(C) = 1 - (0.30 + 0.50) = 0.20$. The conditional probabilities are $P(E \mid A) = 0.6$,$P(E \mid B) = 0.3$,and $P(E \mid C) = 0.1$. Using Bayes' Theorem,the probability $P(C \mid E)$ is given by: $P(C \mid E) = \frac{P(C) P(E \mid C)}{P(A) P(E \mid A) + P(B) P(E \mid B) + P(C) P(E \mid C)}$ Substituting the values: $P(C \mid E) = \frac{0.20 	imes 0.1}{(0.30 	imes 0.6) + (0.50 	imes 0.3) + (0.20 	imes 0.1)}$ $P(C \mid E) = \frac{0.02}{0.18 + 0.15 + 0.02} = \frac{0.02}{0.35} = \frac{2}{35}$.

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