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?
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?
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1 Answer
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41. Let, event that the driver is scooter driver
event that the driver is car driver
event that the driver is truck driver
event the person meets with accident
Total number of drivers
(driver is a scooter driver)
(driver is a car driver)
(driver is a truck driver)
(scooter driver met with an accident)
(car driver met with an accident)
(truck driver met with an accident)
The probability that the driver is scooter driver, given that he met with an accident is given by
By Baye’s theorem,
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P (2 obtained on even numbered toss) = k (let)
P (2) =
P (
If x = 0, y = 6, 7, 8, 9, 10
If x = 1, y = 7, 8, 9, 10
If x = 2, y = 8, 9, 10
If x = 3, y = 9, 10
If x = 4, y = 10
If x = 5, y = no possible value
Total possible ways = (5 + 4 + 3 + 2 + 1) * 2
= 30
Required probability
P (2W and 2B) = P (2B, 6W) × P (2W and 2B)
+ P (3B, 5W) × P (2W and 2B)
+ P (4B, 4W) × P (2W and 2B)
+ P (5B, 3W) × P (2W and 2B)
+ P (6B, 2W) × P (2W and 2B)
(15 + 30 + 36 + 30 + 15)
Let probability of tail is
⇒ Probability of getting head =
∴ Probability of getting 2 heads and 1 tail
ax2 + bx + c = 0
D = b2 – 4ac
D = 0
b2 – 4ac = 0
b2 = 4ac
(i) AC = 1, b = 2 (1, 2, 1) is one way
(ii) AC = 4, b = 4
(iii) AC = 9, b = 6, a = 3, c = 3 is one way
1 + 3 + 1 = 5 way
Required probability =
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