An insurance company selected 2000 drivers at random (i.e., without an

1.	An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained is given in the following table.Age of drivers (In years)Accidents in one yearMore than 3 accidents012318-29440160110613530-50505125602218over 503604535159Find the probabilities of the following events for a driver chosen at random from the city: i) The driver being in the age group 18-29 years and having exactly 3 accidents in one year.ii) The driver being in the age group of 30 – 50 years and having one or more accidents in a year.iii) Having no accidents in the year.
Answer» i) Total number of accidents = (440 + 160 + 110 + 61 + 35 + 505 + 125 + 60 + 22 + 18 + 360 + 45 + 35 + 15 + 9) = 2000 Event: The driver being in the age group (18 – 29) years and having exactly 3 accidents = 61 P(E) = \(\frac {No.\,of\, favourables}{Total\, outcomes}\) = 61/2000 ii) Favourable outcomes = 125 + 60 + 22 + 18 = 225 Total outcomes = 2000 P(E) = 225/2000 = 9/80 iii) Favourable outcomes = 440 + 505 + 360 = 1305 Total outcomes = 2000 P(E) = 1305/2000 = 261/400

An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained is given in the following table.Age of drivers (In years)Accidents in one yearMore than 3 accidents012318-29440160110613530-50505125602218over 503604535159Find the probabilities of the following events for a driver chosen at random from the city: i) The driver being in the age group 18-29 years and having exactly 3 accidents in one year.ii) The driver being in the age group of 30 – 50 years and having one or more accidents in a year.iii) Having no accidents in the year.

Answer»

i) Total number of accidents = (440 + 160 + 110 + 61 + 35 + 505 + 125 + 60 + 22 + 18 + 360 + 45 + 35 + 15 + 9) = 2000

Event: The driver being in the age group (18 – 29) years and having exactly 3 accidents = 61

P(E) = \(\frac {No.\,of\, favourables}{Total\, outcomes}\) = 61/2000

ii) Favourable outcomes = 125 + 60 + 22 + 18 = 225

Total outcomes = 2000

P(E) = 225/2000 = 9/80

iii) Favourable outcomes = 440 + 505 + 360 = 1305

Total outcomes = 2000

P(E) = 1305/2000 = 261/400