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An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively.

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प्रश्न

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

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उत्तर

Let E1: be the scooter driver.

E2: The car is a driver.

E3:  Is a truck driver.

E: The insured driver gets into an accident

Driver = 2000 + 4000 + 6000 = 12000

P(E1) = `2000/12000 = 1/6`,

P(E2) = `4000/12000 = 1/3`,

P(E3) = `6000/12000 = 1/2`

P(E|E1) − 0.01, P(E|E2) = 0.03, P(E|E3) = 0.15

The probability that the accident driver is a scooter driver = P(E1|E)

By Bayes' theorem

P(E1|E) = `(P(E_1) xx P(E|E_1))/(P(E_1) xx P(E|E_1) + P(E_2) xx P(E|E_2) + P(E_3) xx P(E|E_3)`

= `(1/6 xx (0.01))/(0.01/6 + 0.03/3 + 0.15/2)`

= `0.01/(0.01 + 0.06 + 0.45)`

= `0.01/0.52`

= `1/52`

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अध्याय 13: Probability - Exercise 13.3 [पृष्ठ ५५६]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 13 Probability
Exercise 13.3 | Q 7 | पृष्ठ ५५६

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