Question

# If $5$ of a company’s $10$ delivery trucks do not meet emission standards and $3$ of them are chosen for inspection, then what is the probability that none of the trucks chosen will meet emission standards?A. $\dfrac{1}{8}$B. $\dfrac{3}{8}$C. $\dfrac{1}{{12}}$D. $\dfrac{1}{4}$

Hint: To find the required probability, first we will consider the event $E$ that none of trucks chosen will meet emission standards. We will find the required probability by using the definition. That is, required probability $= \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$ where $n\left( E \right)$ is the number of favourable (desired) outcomes and $n\left( S \right)$ is the number of total outcomes.

In this problem, it is given that $5$ trucks out of $10$ delivery trucks do not meet emission standard. Also given that $3$ trucks are selected for inspection. Therefore, we can write the given information in the following way:
Total number of delivery trucks $= 10$
Total number of trucks which do not meet emission standard $= 5$
Total number of selected trucks $= 3$
Let us consider the event $E$ that none of trucks chosen will meet emission standards. That is, all selected trucks will not meet emission standards. There are $5$ trucks which do not meet emission standard and we need to select $3$ trucks for inspection. We know that the number of ways of selecting $3$ trucks out of $5$ trucks is given by ${}^5{C_3}$. So, we can say that the total number of favourable outcomes is ${}^5{C_3}$. That is, $n\left( E \right) = {}^5{C_3}$.
There are total $10$ trucks and we need to select $3$ trucks for inspection. We know that the number of ways of selecting $3$ trucks out of $10$ trucks is given by ${}^{10}{C_3}$. So, we can say that the total number of outcomes is ${}^{10}{C_3}$. That is, $n\left( S \right) = {}^{10}{C_3}$.
Now we are going to find the probability of an event $E$ by using the definition. That is,
$P$( none of trucks chosen will meet emission standards ) $= P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
$\Rightarrow P\left( E \right) = \dfrac{{{}^5{C_3}}}{{{}^{10}{C_3}}}$
Let us find ${}^5{C_3}$ and ${}^{10}{C_3}$ by using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Therefore, we get
${}^5{C_3} = \dfrac{{5!}}{{\left( {3!} \right) \times \left( {5 - 3} \right)!}} = \dfrac{{1 \times 2 \times 3 \times 4 \times 5}}{{\left( {1 \times 2 \times 3} \right) \times 2!}} = \dfrac{{4 \times 5}}{{1 \times 2}} = \dfrac{{20}}{2} = 10$ and
${}^{10}{C_3} = \dfrac{{10!}}{{\left( {3!} \right) \times \left( {10 - 3} \right)!}} = \dfrac{{10!}}{{\left( {3!} \right) \times 7!}} = \dfrac{{\left( {7!} \right) \times \left( {8 \times 9 \times 10} \right)}}{{\left( {1 \times 2 \times 3} \right) \times 7!}} = 120$
Let us substitute these values in $P\left( E \right) = \dfrac{{{}^5{C_3}}}{{{}^{10}{C_3}}}$. Therefore, we get $P\left( E \right) = \dfrac{{10}}{{120}} = \dfrac{1}{{12}}$. Therefore, the probability that none of the trucks chosen will meet emission standards is $\dfrac{1}{{12}}$.
Therefore, option C is correct.

Note: For any event $A$, we can write $0 \leqslant P\left( A \right) \leqslant 1$ where $P\left( A \right)$ is the probability of event $A$. The sum of probabilities of all possible outcomes is always $1$. These are the properties of basic probability theory.