Question

The ratio of the number of trucks along a highway on which a petrol pump is located, to the number of cars running along highways is 3:2. It is known that an average of one truck in thirty trucks and two cars in fifty cars stop at the petrol pump to be filled up with the fuel. If a vehicle stops at the petrol pump to be filled up with the fuel, find the probability that it is a car,
A. \[\dfrac{4}{9}\]
B. \[\dfrac{9}{{250}}\]
C. \[\dfrac{3}{5}\]
D. \[\dfrac{1}{{30}}\]

Answer 1

Hint: To solve the given question for probability you have to understand the concept of probability that is probability is the ratio of the favorable outcome to the total outcome for any event. For example probability of coming head for tossing a coin would be half, because here favorable outcome for this event id one and total outcome is two that is coin can come as head and tail both.

Formulae Used:
\[
   \Rightarrow P(Z) = P(CZ \cup T{Z_1}) \\
   \Rightarrow P(Z) = P(CZ) + P(T{Z_1}) \\
 \]

Complete step by step answer:
The given question need to be solved for the probability of car to be stopped in petrol pump, for which we are going to solve with conditional probability:
On solving we get:
Let stopping a car at petrol pump=C
Let stopping a truck at petrol pump=T
Ratio of trucks to car=3:2
Number of car=2x
Number of trucks=3x
Total number of cars and trucks=2x+3x=5x
It is known that for every 30 truck, one truck stops and for every 50 cars 2 cars stops
The stopped vehicle is car then probability \[P(C) = \dfrac{{2x}}{{5x}} = \dfrac{2}{5}\]
The stopped vehicle is truck then probability \[P(C) = \dfrac{{3x}}{{5x}} = \dfrac{3}{5}\]
Vehicle stops for fuel for car \[P\left( {\dfrac{Z}{C}} \right) = \dfrac{2}{{50}}\](C= maximum number of vehicle that is given 50 for car)
Vehicle stops for fuel for truck \[P\left( {\dfrac{Z}{T}} \right) = \dfrac{1}{{30}}\](T=maximum number of vehicle that is given 30 for truck)
\[
   \Rightarrow P(Z) = P(CZ \cup T{Z_1}) \\
   \Rightarrow P(Z) = P(CZ) + P(T{Z_1}) \\
   \Rightarrow P(Z) = \dfrac{2}{5} \times \dfrac{2}{{50}} + \dfrac{3}{5} \times \dfrac{1}{{30}} = \dfrac{4}{{250}} + \dfrac{3}{{150}} = \dfrac{{12 + 15}}{{750}} = \dfrac{{27}}{{750}} = \dfrac{9}{{250}} \\
   \Rightarrow P\left( {\dfrac{C}{Z}} \right) = \dfrac{{P(CZ)}}{{P(Z)}} = \dfrac{{\dfrac{4}{{250}}}}{{\dfrac{9}{{250}}}} = \dfrac{4}{9} \\
 \]

So, the correct answer is Option A.

Note: Here for this question we have used conditional probability which explains for the question in which condition are given for an event, steps used for the question are needed to be solved for getting the answer, you have to assume the variables then after using conditional probability we can obtain our required answer.