Question

At a traffic light on 28th April, out of 310 vehicles which crossed
the light, 200 were cars, 60 were two wheelers & 50 were autos
18 were fined for jumping the red light or not wearing of belt or
helmet, 5 were fined for using car with odd number, four wereleft
after giving warning. What is the probability that
1) Acar is chosen & it bears even number.
1) A fine was given
​

Answer 1

Given:

No. of vehicles crossed the light = 310

No. of cars = 200

No. of two wheelers = 60

No. of autos = 50

No. of people fined for not wearing helmet or belt or for jumping red light = 18

No. of people fined for using odd number car = 5

No. of people given warning = 4

To find:

Probability that a car is chosen and it has an even number.

Probability that a fine was given.

Solution:

Total no. of cars that crossed the light = 200

No. of cars with odd number = 5

No. of cars with even number = $200-5=195$

Probability of being a car and it is even = $\frac{200}{310}$ × $\frac{195}{200}$ = $\frac{39}{62}$

Total no. of vehicles that were fined = 18 + 5 = 23

Probability that a vehicle was fined = $\frac{23}{310}$

The probability that a car is chosen and it bears even number is $\frac{39}{62}$

The probability that a vehicle was fined is $\frac{23}{310}$