Question

There are 5 duplicates and 10 original items in an automobile shop and 3 items are bought at random by a customer. The probability that none of the items is duplicate is:
A. ${\displaystyle \frac{20}{91}}$
B. ${\displaystyle \frac{22}{91}}$
C.${\displaystyle \frac{24}{91}}$
D.${\displaystyle \frac{89}{91}}$

Answer 1

Hint: The number of ways of selecting $r$ items from $n$ different items is determined by the formula ${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$. Use this formula to find out the number of favorable ways and total number of ways of the given event. The probability is the ratio of them i.e. $\text{Probability}={\displaystyle \frac{\text{No}\text{. of favorable ways}}{\text{Total number of ways}}}$.

Complete step by step answer:
According to the question, there are $5$ duplicates and $10$ original items in an automobile shop.
So the total number of items is $5+10$ i.e. $15$. And $3$ items are bought at random from them.
We know that the number of ways of selecting $r$ items from $n$ different items is determined by the formula ${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$.
Thus the number of ways of selecting $3$ items from total $15$ items is given as:
$$\begin{gathered} \Rightarrow \text{Total number of ways}{=}^{15}{C}_3={\displaystyle \frac{15!}{3!12!}}={\displaystyle \frac{15\times 14\times 13\times 12!}{6\times 12!}} \\ \Rightarrow \text{Total number of ways}=455.....\text{(1)} \end{gathered}$$
Again from the question, all the selected items should be original and not duplicate. Hence, the number of ways of selecting $3$ items from $10$ original items is given as:
$$\begin{gathered} \Rightarrow \text{ No}\text{. of favorable ways}{=}^{10}{C}_3={\displaystyle \frac{10!}{3!7!}}={\displaystyle \frac{10\times 9\times 8\times 7!}{6\times 7!}} \\ \Rightarrow \text{No}\text{. of favorable ways}=120.....\text{(2)} \end{gathered}$$
Further, we know that probability of an event is calculated by the formula:
$\Rightarrow \text{Probability}={\displaystyle \frac{\text{No}\text{. of favorable ways}}{\text{Total number of ways}}}$
Therefore, applying this formula and putting the values from equation (1) and (2), we’ll get:
$$\begin{gathered} \Rightarrow \text{Probability}={\displaystyle \frac{120}{455}}={\displaystyle \frac{24\times 5}{91\times 5}} \\ \Rightarrow \text{Probability}={\displaystyle \frac{24}{91}} \end{gathered}$$

$\therefore$ The required probability of selecting $3$ original items from the given items is ${\displaystyle \frac{24}{91}}$. Hence, option (C) is the correct option.

Note:
The formula used for combination i.e. ${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$ is just for the number of ways of selecting the items and not for arranging them.
The number of ways of selecting and arranging $r$ items from $n$ different items is given by the formula of permutation and this formula is:
 ${\Rightarrow }^n{P}_r={\displaystyle \frac{n!}{\left(n-r\right)!}}$