The probability that a person goes to college | vedclass

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Question

(C) Let $C$,$B$,and $T$ be the events that the person travels by car,bus,and train respectively. Let $L$ be the event that the person reaches college

Vedclass · Accepted Answer

$\frac{5}{29}$

Vedclass · Answer

(C) Let $C$,$B$,and $T$ be the events that the person travels by car,bus,and train respectively. Let $L$ be the event that the person reaches college late,and $L'$ be the event that the person reaches college in time. Given probabilities: $P(C) = \frac{1}{5}$,$P(B) = \frac{2}{5}$,$P(T) = \frac{3}{5}$. Probabilities of being late: $P(L|C) = \frac{2}{7}$,$P(L|B) = \frac{4}{7}$,$P(L|T) = \frac{1}{7}$. Probabilities of being on time: $P(L'|C) = 1 - \frac{2}{7} = \frac{5}{7}$,$P(L'|B) = 1 - \frac{4}{7} = \frac{3}{7}$,$P(L'|T) = 1 - \frac{1}{7} = \frac{6}{7}$. Using Bayes' Theorem,the probability that he traveled by car given he reached in time is: $P(C|L') = \frac{P(L'|C)P(C)}{P(L'|C)P(C) + P(L'|B)P(B) + P(L'|T)P(T)}$ $P(C|L') = \frac{(\frac{5}{7} 	imes \frac{1}{5})}{(\frac{5}{7} 	imes \frac{1}{5}) + (\frac{3}{7} 	imes \frac{2}{5}) + (\frac{6}{7} 	imes \frac{3}{5})}$ $P(C|L') = \frac{\frac{5}{35}}{\frac{5}{35} + \frac{6}{35} + \frac{18}{35}} = \frac{5}{5 + 6 + 18} = \frac{5}{29}$.

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