Question

On a particular day a policeman observed vehicles for a speed check. The frequency table shows the speed of 160 vehicles that pass a radar speed check on dual carriageways.

Speed (km/h)	20-29	30-39	40-49	50-59	60-69	70 and above
Number of vehicles	14	23	28	35	52	8

Find the probability that the speed of a vehicle selected at random is faster than 69km/h.

Answer 1

For solving this problem we use the general formula of probability that is if $E$ is the event, for which we need to find the probability then the formula is given as $P\left( E \right)=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}$. We need to find the number of possible vehicles that have a speed of more than 69km/h and total possible vehicles and substitute in the above formula to get the required answer.
Complete step-by-step solution
Let us assume that $E$ be the event of getting a vehicle having a speed greater than 69km/h.
In the table given we can see that there are ‘8’ vehicles that are having speed more than 69km/h.
When one vehicle is selected at random, for the possible outcomes of getting a vehicle having a speed greater than 69km/h, we need to select that vehicle from those ‘8’ vehicles having speed 70km/h and above.
Therefore the possible outcomes of the event $E$ are selecting one vehicle from ‘8’ vehicles that are equal to ${}^{8}{{C}_{1}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
By applying the above formula we get

& \Rightarrow \text{possible outcomes = }{}^{8}{{C}_{1}} \\\ & \Rightarrow \text{possible outcomes =}\dfrac{8!}{1!\times 7!}=8 \\\ \end{aligned}$$ We are given that there are a total of 160 vehicles, when we select one vehicle to random we have total outcomes as $${}^{160}{{C}_{1}}$$. By applying the combinations formula we get $$\begin{aligned} & \Rightarrow \text{total outcomes = }{}^{160}{{C}_{1}} \\\ & \Rightarrow \text{total outcomes =}\dfrac{160!}{1!\times 159!}=160 \\\ \end{aligned}$$ We know that probability of event as $$P\left( E \right)=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}$$ By substituting the required values in above equation we get $$\begin{aligned} & \Rightarrow P\left( E \right)=\dfrac{8}{160} \\\ & \Rightarrow P\left( E \right)=\dfrac{1}{20} \\\ \end{aligned}$$ **Therefore the answer is $$\dfrac{1}{20}$$.** **Note:** Students may make mistakes in taking the possible outcomes. In the question, we are asked to find the probability of a vehicle having a speed of more than 69km/h. But students may also consider the vehicles having a speed of 69km/h and take the total possible outcomes as $$52+8=60$$ which will be wrong. We are asked for more than 69km/h, we are not asked for equal to or more than 69km/h. This is the only point that needs to be taken care of.