Questions & Answers

Question

Answers

A. $\dfrac{1}{8}$

B. $\dfrac{3}{8}$

C. $\dfrac{1}{{12}}$

D. $\dfrac{1}{4}$

Answer
Verified

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 $\dfrac{1}{{12}}$.