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An insurance company insured 1000 scooter drivers, 2000 car drivers and 4000 truck drivers. The probability of accidents by scooter, car and truck drivers are 0.02, 0.05 and 0.03 respectively. - Mathematics

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Question

An insurance company insured 1000 scooter drivers, 2000 car drivers and 4000 truck drivers. The probability of accidents by scooter, car and truck drivers are 0.02, 0.05 and 0.03 respectively. If one of the insured persons meets with an accident, find the probability that he is a truck driver.

Sum
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Solution

Let, the probability of being a scooter driver

P(A) = `1000/7000`

= `1/7`

Similarly, P(B) (Probability of being car drivers) 

= `2000/7000`

= `2/7`

P(C) (Probability of being a truck driver)

= `4000/7000`

= `4/7`

Let E be the event that drivers meet with an accident:

So, P(E/A) = `2/100`

P(E/B) = `5/100`

P(E/C) = `3/100`

Now,

P(C/E)= `(P(C)  .  P(E/C))/(P(A)  .  P(E/A) + P(B)  .  P(E/B) + P(C)  .  P(E/C))`

= `(4/7 xx 3/100)/(1/7 xx 2/100 + 2/7 xx 5/100 + 4/7xx 3/100)`

= `(12/700)/(2/700 + 10/700 + 12/700)`

= `(12/700)/(24/700)`

= `1/2`

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