A candidate has to go to the examination centre using only one means of transportation out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car respectively are 52,51,52. The probabilities of reaching late are 31,41,11...
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The probabilities that the candidate reaches late are 31 if by bus, 41 if by scooter, and 51 --- *note: the exact "car" late-probability could not be reliably extracted; based on the solution working, P(late∣bus)=51, P(late∣scooter)=31, P(late∣car)=41 is what is actually used.* Given that the candidate reached late, the probability that the candidate travelled by bus is:
Solutions · 1
Official solution
Let B,S,C be the events of travelling by bus, scooter, car; L: arriving late.
\[P(B)=\frac25,\ P(S)=\frac15,\ P(C)=\frac25;\qquad P(L\mid B)=\frac15,\ P(L\mid S)=\frac13,\ P(L\mid C)=\frac14\]
\[P(B\mid L)=\frac{P(B)P(L\mid B)}{P(B)P(L\mid B)+P(S)P(L\mid S)+P(C)P(L\mid C)} = \frac{\left(\frac25\right)\left(\frac15\right)}{\left(\frac25\right)\left(\frac15\right)+\left(\frac15\right)\left(\frac13\right)+\left(\frac25\right)\left(\frac14\right)}\]
\[=\frac{2/25}{2/25+1/15+1/10} = \left(\frac{2}{25}\right)\left(\frac{300}{24+20+30}\right)=\frac{12}{37}\]