A doctor is to visit a patient. From past exp | vedclass

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(A) Let $E$ be the event that the doctor arrives late. Let $T_1, T_2, T_3,$ and $T_4$ be the events that the doctor comes by train,bus,scooter,and oth

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $E$ be the event that the doctor arrives late. Let $T_1, T_2, T_3,$ and $T_4$ be the events that the doctor comes by train,bus,scooter,and other means of transport,respectively.Given probabilities:$P(T_1) = \frac{3}{10}, P(T_2) = \frac{1}{5}, P(T_3) = \frac{1}{10}, P(T_4) = \frac{2}{5}$.Conditional probabilities of being late:$P(E|T_1) = \frac{1}{4}, P(E|T_2) = \frac{1}{3}, P(E|T_3) = \frac{1}{12}, P(E|T_4) = 0$.Using Bayes' Theorem,the probability that he comes by train given he is late is:$P(T_1|E) = \frac{P(T_1)P(E|T_1)}{\sum_{i=1}^{4} P(T_i)P(E|T_i)}$Denominator calculation:$P(E) = (\frac{3}{10} 	imes \frac{1}{4}) + (\frac{1}{5} 	imes \frac{1}{3}) + (\frac{1}{10} 	imes \frac{1}{12}) + (\frac{2}{5} 	imes 0)$$P(E) = \frac{3}{40} + \frac{1}{15} + \frac{1}{120} + 0 = \frac{9 + 8 + 1}{120} = \frac{18}{120} = \frac{3}{20}$.Numerator calculation:$P(T_1)P(E|T_1) = \frac{3}{10} 	imes \frac{1}{4} = \frac{3}{40}$.Result:$P(T_1|E) = \frac{3/40}{3/20} = \frac{3}{40} 	imes \frac{20}{3} = \frac{1}{2}$.

Box	White	Green	Red	Probability
$A$	$1$	$2$	$3$	$\frac{1}{2}$
$B$	$2$	$3$	$1$	$\frac{1}{3}$
$C$	$3$	$1$	$2$	$\frac{1}{6}$

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