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 the emission standards?
A) $\dfrac{1}{8}$
B) $\dfrac{3}{8}$
C) $\dfrac{1}{{12}}$
D) $\dfrac{1}{4}$

Answer 1

We can find the number of ways of selecting 3 delivery trucks from 10 trucks using combinations. This will be the number of the total possible outcomes. Then we can find the number of ways of selecting 3 trucks from the 5 trucks that do not meet the emission standards. This will be the number of favourable outcomes. Then we can find the required probability by dividing the number of favourable outcomes by the number of the total possible outcomes.

Complete step by step solution:
We are given that there are 10 delivery trucks and out of which 5 do not meet emission standards.
We need to select 3 trucks.
So, the number of ways of selecting 3 trucks from the total of 10 trucks is given by ${}^{10}{C_3}$.
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
On simplifying the combination, we get
$\Rightarrow {}^{10}{C_3} = \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}}$
So, we have,
$\Rightarrow {}^{10}{C_3} = \dfrac{{10!}}{{3! \times 7!}}$
On expanding the factorials, we get
$\Rightarrow {}^{10}{C_3} = \dfrac{{10 \times 9 \times 8 \times 7!}}{{3 \times 2 \times 7!}}$
On cancelling the common terms, we get
$\Rightarrow {}^{10}{C_3} = 10 \times 3 \times 4$
On multiplication we get
$\Rightarrow {}^{10}{C_3} = 120$
Therefore, the number of ways to select 3 trucks is 120.
Now the number of ways of selecting 3 trucks from the 5 trucks that do not meet the emission standard is given by ${}^5{C_3}$.
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
On simplifying the combination, we get
$\Rightarrow {}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}}$
So, we have,
$\Rightarrow {}^5{C_3} = \dfrac{{5!}}{{3! \times 2!}}$
On expanding the factorials, we get
$\Rightarrow {}^5{C_3} = \dfrac{{5 \times 4 \times 3!}}{{3! \times 2}}$
On cancelling the common terms, we get
$\Rightarrow {}^5{C_3} = 10$
Therefore, the number of ways of selecting 3 trucks which are defective is 10.
Now we need to find the probability that 3 trucks are chosen and none of the trucks chosen will meet the emission standards.
We know that the probability of an event is given by the number of favourable outcomes divided by the number of total numbers of possible outcomes.
$\Rightarrow P = \dfrac{{no.\,of\,favourable\,outcome}}{{total\,no.\,of\,outcome}}$
Here the number of the total possible outcomes is the number of ways selecting 3 trucks which is 120 and the number of favourable outcomes is the number of ways selecting 3 trucks that are defective which is 10.
$\Rightarrow P = \dfrac{{10}}{{120}}$
On simplification, we get
$\Rightarrow P = \dfrac{1}{{12}}$
Therefore, the required probability is $\dfrac{1}{{12}}$.

So, the correct answer is option C.

Note:
Alternate method to solve this problem is given by,
We need to find the probability that 3 trucks are chosen and none of the trucks chosen will meet the emission standards.
We know that the probability of an event is given by the number of favourable outcomes divided by the number of total numbers of possible outcomes.
$\Rightarrow P = \dfrac{{no.\,of\,favourable\,outcome}}{{total\,no.\,of\,outcome}}$
Here the number of the total possible outcomes is the number of ways selecting 3 trucks from 10 truck which is given by ${}^{10}{C_3}$ and the number of favourable outcomes is the number of ways selecting 3 trucks from 5 trucks that are defective which is given by ${}^5{C_3}$
$\Rightarrow P = \dfrac{{{}^5{C_3}}}{{{}^{10}{C_3}}}$
On expanding the combinations using ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow P = \dfrac{{\dfrac{{5!}}{{3! \times 2!}}}}{{\dfrac{{10!}}{{3! \times 7!}}}}$
On rearranging and cancelling the common terms, we get
$\Rightarrow P = \dfrac{{5! \times 7!}}{{10! \times 2!}}$
On expanding the factorials, we get
$\Rightarrow P = \dfrac{{5 \times 4 \times 3 \times 2! \times 7!}}{{10 \times 9 \times 8 \times 7! \times 2!}}$
On cancelling the common terms, we get
$\Rightarrow P = \dfrac{{1 \times 1 \times 1}}{{2 \times 3 \times 2}}$
On simplification we get
$\Rightarrow P = \dfrac{1}{{12}}$
Therefore, the required probability is $\dfrac{1}{{12}}$.