41. An insurance company insured 2000 scooter driver, 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?

0 3 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    41. Let, E1: event that the driver is scooter driver

    E2: event that the driver is car driver

    E3: event that the driver is truck driver

    A: event the person meets with accident

    Total number of drivers =2000+4000+6000=12000

    P(E1)=P (driver is a scooter driver) =200012000=16

    P(E2)=P (driver is a car driver) =400012000=13

    P(E3)=P (driver is a truck driver) =600012000=12

    P(A/E1)=P (scooter driver met with an accident) =0.01=1100

    P(A/E2)=P (car driver met with an accident) =0.03=3100

    P(A/E3)=P (truck driver met with an accident) =0.15=15100

    The probability that the driver is scooter driver, given that he met with an accident is given by P(E1/A)

    By Baye’s theorem,

    P(E1/A)=P(E1).P(A/E1)P(E1).P(A/E1)+P(E2).P(A/E2)+P(E3).P(A/E3)

    =16×110016×1100+13×3100+12×15100

    =16001100(16+1+152)

    =16001100(1+6+456)

    =16001100×526

    =1600×60052=152

     

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A
alok kumar singh

P (2 obtained on even numbered toss) = k (let)

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A
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If x = 0, y = 6, 7, 8, 9, 10

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A
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A
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Let probability of tail is   1 3

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V
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ax2 + bx + c = 0

D = b2 – 4ac

D = 0

b2 – 4ac = 0

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