Q.6. A candidate has to go to the examination centre to appear in an examination. The candidate uses

Q: What is the correct answer for JEE Main Mathematics Q6?

The correct answer is B. Sol. $P(Bus) = P(B) = \frac{2}{5}$ $P(Late|Bus) = P(L|B) = \frac{1}{5}$ $P(Scooter) = P(S) = \frac{1}{5}$ $P(Late|Scooter) = P(L|S) = \frac{1}{3}$ $P(Car) = P(C) = \frac{2}{5}$ $P(Late|Car) = P(L|C) = \frac{1}{4}$ $P(L) = P(L|B)P(B) + P(L|S)P(S) + P(L|C)P(C)$ $P(L) = \frac{1}{5} \times \frac{2}{5} + \frac{1}{3} \times \frac{1}{5} + \frac{1}{4} \times \frac{2}{5} = \frac{2}{25} + \frac{1}{15} + \fr

Q: Q.6. A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The…

The correct answer is B. Sol. $P(Bus) = P(B) = \frac{2}{5}$ $P(Late|Bus) = P(L|B) = \frac{1}{5}$ $P(Scooter) = P(S) = \frac{1}{5}$ $P(Late|Scooter) = P(L|S) = \frac{1}{3}$ $P(Car) = P(C) = \frac{2}{5}$ $P(Late|Car) = P(L|C) = \frac{1}{4}$ $P(L) = P(L|B)P(B) + P(L|S)P(S) + P(L|C)P(C)$ $P(L) = \frac{1}{5} \times \frac{2}{5} + \frac{1}{3} \times \frac{1}{5} + \frac{1}{4} \times \frac{2}{5} = \frac{2}{25} + \frac{1}{15} + \fr

Q: What are the options for this JEE Main Mathematics question?

A. 11/37 · B. 12/37 · C. 13/37 · D. 14/37

Question

Q.6. A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are  and . The probabilities that the candidate reaches late at the examination centre are  and  if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is :

FirstInTest · Accepted Answer

Sol. $P(Bus) = P(B) = \frac{2}{5}$ $P(Late|Bus) = P(L|B) = \frac{1}{5}$ $P(Scooter) = P(S) = \frac{1}{5}$ $P(Late|Scooter) = P(L|S) = \frac{1}{3}$ $P(Car) = P(C) = \frac{2}{5}$ $P(Late|Car) = P(L|C) = \frac{1}{4}$ $P(L) = P(L|B)P(B) + P(L|S)P(S) + P(L|C)P(C)$ $P(L) = \frac{1}{5} 	imes \frac{2}{5} + \frac{1}{3} 	imes \frac{1}{5} + \frac{1}{4} 	imes \frac{2}{5} = \frac{2}{25} + \frac{1}{15} + \frac{2}{20} = \frac{2}{25} + \frac{1}{15} + \frac{1}{10}$ $P(L) = \frac{12+10+15}{150} = \frac{37}{150}$ We need to find $P(B|L) = \frac{P(L|B)P(B)}{P(L)} = \frac{\frac{1}{5} 	imes \frac{2}{5}}{\frac{37}{150}} = \frac{2}{25} 	imes \frac{150}{37} = \frac{2 	imes 6}{37} = \frac{12}{37}$

Answer

11/37

Answer

12/37

Answer

13/37

Answer

14/37

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