An insurance company selected $2000$ drivers at random in a particular city to find a relationship between age and accidents. The data obtained are given in the following table:
Age of drivers (in years) $0$ accidents $1$ accident $2$ accidents $3$ accidents Over $3$ accidents $18-29$ $440$ $160$ $110$ $61$ $35$ $30-50$ $505$ $125$ $60$ $22$ $18$ Above $50$ $360$ $45$ $35$ $15$ $9$
Find the probabilities of the following events for a driver chosen at random from the city:
$(i)$ Being $18-29$ years of age and having exactly $3$ accidents in one year.
$(ii)$ Being $30-50$ years of age and having one or more accidents in a year.
$(iii)$ Having no accidents in one year.
| Age of drivers (in years) | $0$ accidents | $1$ accident | $2$ accidents | $3$ accidents | Over $3$ accidents |
| $18-29$ | $440$ | $160$ | $110$ | $61$ | $35$ |
| $30-50$ | $505$ | $125$ | $60$ | $22$ | $18$ |
| Above $50$ | $360$ | $45$ | $35$ | $15$ | $9$ |
Find the probabilities of the following events for a driver chosen at random from the city:
$(i)$ Being $18-29$ years of age and having exactly $3$ accidents in one year.
$(ii)$ Being $30-50$ years of age and having one or more accidents in a year.
$(iii)$ Having no accidents in one year.
(N/A) Total number of drivers $= 2000$.
$(i)$ The number of drivers who are $18-29$ years old and have exactly $3$ accidents in one year is $61$.
So,$P$ (driver is $18-29$ years old with exactly $3$ accidents) $= \frac{61}{2000} = 0.0305$.
$(ii)$ The number of drivers $30-50$ years of age and having one or more accidents in one year $= 125 + 60 + 22 + 18 = 225$.
So,$P$ (driver is $30-50$ years of age and having one or more accidents) $= \frac{225}{2000} = 0.1125$.
$(iii)$ The number of drivers having no accidents in one year $= 440 + 505 + 360 = 1305$.
Therefore,$P$ (drivers with no accident) $= \frac{1305}{2000} = 0.6525$.
$(i)$ The number of drivers who are $18-29$ years old and have exactly $3$ accidents in one year is $61$.
So,$P$ (driver is $18-29$ years old with exactly $3$ accidents) $= \frac{61}{2000} = 0.0305$.
$(ii)$ The number of drivers $30-50$ years of age and having one or more accidents in one year $= 125 + 60 + 22 + 18 = 225$.
So,$P$ (driver is $30-50$ years of age and having one or more accidents) $= \frac{225}{2000} = 0.1125$.
$(iii)$ The number of drivers having no accidents in one year $= 440 + 505 + 360 = 1305$.
Therefore,$P$ (drivers with no accident) $= \frac{1305}{2000} = 0.6525$.
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