In a group of 400 people,160 are smokers and | vedclass

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Question

(C) Let the events be defined as follows:$A$: The person chosen is a smoker and non-vegetarian.$B$: The person chosen is a smoker and vegetarian.$C$:

Vedclass · Accepted Answer

$\frac{28}{45}$

Vedclass · Answer

(C) Let the events be defined as follows:$A$: The person chosen is a smoker and non-vegetarian.$B$: The person chosen is a smoker and vegetarian.$C$: The person chosen is a non-smoker and vegetarian.$E$: The person chosen has a chest disorder.Given probabilities:$P(A) = \frac{160}{400}$,$P(B) = \frac{100}{400}$,$P(C) = \frac{140}{400}$.Conditional probabilities of having the disorder:$P(E|A) = \frac{35}{100}$,$P(E|B) = \frac{20}{100}$,$P(E|C) = \frac{10}{100}$.Using Bayes' Theorem,we need to find $P(A|E)$:$P(A|E) = \frac{P(A) \cdot P(E|A)}{P(A) \cdot P(E|A) + P(B) \cdot P(E|B) + P(C) \cdot P(E|C)}$Substituting the values:$P(A|E) = \frac{\frac{160}{400} 	imes \frac{35}{100}}{\frac{160}{400} 	imes \frac{35}{100} + \frac{100}{400} 	imes \frac{20}{100} + \frac{140}{400} 	imes \frac{10}{100}}$$P(A|E) = \frac{160 	imes 35}{(160 	imes 35) + (100 	imes 20) + (140 	imes 10)}$$P(A|E) = \frac{5600}{5600 + 2000 + 1400} = \frac{5600}{9000} = \frac{56}{90} = \frac{28}{45}$.

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