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A doctor is to visit a patient. It is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 310,15,110\frac{3}{10}, \frac{1}{5}, \frac{1}{10} and 25\frac{2}{5}. The probabilities that he will be late are 14,13\frac{1}{4}, \frac{1}{3} and 112\frac{1}{12}, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he arrives late.

The probability that he comes by bus is:

Solution

✅ Correct Option: 1

By Bayes' theorem, P(late)=(3/10)(1/4)+(1/5)(1/3)+(1/10)(1/12)+(2/5)(0)P(\text{late}) = (3/10)(1/4) + (1/5)(1/3) + (1/10)(1/12) + (2/5)(0). Common denominator 120: 9/120+8/120+1/120=18/120=3/209/120 + 8/120 + 1/120 = 18/120 = 3/20. Then P(bus∣late)=[(1/5)(1/3)]/(3/20)=(1/15)(20/3)=4/9P(\text{bus}|\text{late}) = [(1/5)(1/3)]/(3/20) = (1/15)(20/3) = 4/9.

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