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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Similar Questions
The distance (in $km$) of $40$ engineers from their residence to their place of work were found as follows.
$5$ $3$ $10$ $20$ $25$ $11$ $13$ $7$ $12$ $31$ $19$ $10$ $12$ $17$ $18$ $11$ $32$ $17$ $16$ $2$ $7$ $9$ $7$ $8$ $3$ $5$ $12$ $15$ $18$ $3$ $12$ $14$ $2$ $9$ $6$ $15$ $15$ $7$ $6$ $12$
What is the empirical probability that an engineer lives:
$(i)$ less than $7 \, km$ from her place of work?
$(ii)$ more than or equal to $7 \, km$ from her place of work?
$(iii)$ within $\frac{1}{2} \, km$ from her place of work?
| $5$ | $3$ | $10$ | $20$ | $25$ | $11$ | $13$ | $7$ | $12$ | $31$ |
| $19$ | $10$ | $12$ | $17$ | $18$ | $11$ | $32$ | $17$ | $16$ | $2$ |
| $7$ | $9$ | $7$ | $8$ | $3$ | $5$ | $12$ | $15$ | $18$ | $3$ |
| $12$ | $14$ | $2$ | $9$ | $6$ | $15$ | $15$ | $7$ | $6$ | $12$ |
What is the empirical probability that an engineer lives:
$(i)$ less than $7 \, km$ from her place of work?
$(ii)$ more than or equal to $7 \, km$ from her place of work?
$(iii)$ within $\frac{1}{2} \, km$ from her place of work?
Medium
View SolutionConsider the frequency distribution table which gives the weights of $38$ students of a class.
Weights (in $kg$) Number of students $31-35$ $9$ $36-40$ $5$ $41-45$ $14$ $46-50$ $3$ $51-55$ $1$ $56-60$ $2$ $61-65$ $2$ $66-70$ $1$ $71-75$ $1$ Total $38$
$(i)$ Find the probability that the weight of a student in the class lies in the interval $46-50 \, kg$.
$(ii)$ Give two events in this context,one having probability $0$ and the other having probability $1$.
| Weights (in $kg$) | Number of students |
| $31-35$ | $9$ |
| $36-40$ | $5$ |
| $41-45$ | $14$ |
| $46-50$ | $3$ |
| $51-55$ | $1$ |
| $56-60$ | $2$ |
| $61-65$ | $2$ |
| $66-70$ | $1$ |
| $71-75$ | $1$ |
| Total | $38$ |
$(i)$ Find the probability that the weight of a student in the class lies in the interval $46-50 \, kg$.
$(ii)$ Give two events in this context,one having probability $0$ and the other having probability $1$.
Medium
View SolutionThe percentage of marks obtained by a student in the monthly unit tests are given below:
Unit test $I$ $II$ $III$ $IV$ $V$ Percentage of marks obtained $69$ $71$ $73$ $68$ $74$
Based on this data,find the probability that the student gets more than $70\%$ marks in a unit test.
| Unit test | $I$ | $II$ | $III$ | $IV$ | $V$ |
| Percentage of marks obtained | $69$ | $71$ | $73$ | $68$ | $74$ |
Based on this data,find the probability that the student gets more than $70\%$ marks in a unit test.
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View SolutionAsk all the students in your class to write a $3-$digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by $3$? Remember that a number is divisible by $3$,if the sum of its digits is divisible by $3$.
Medium
View SolutionThe record of a weather station shows that out of the past $250$ consecutive days,its weather forecasts were correct $175$ times.
$(i)$ What is the probability that on a given day it was correct?
$(ii)$ What is the probability that it was not correct on a given day?
$(i)$ What is the probability that on a given day it was correct?
$(ii)$ What is the probability that it was not correct on a given day?
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