Answer
436.8k+ views
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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)