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. $\dfrac{{20}}{{91}}$
B. $\dfrac{{22}}{{91}}$
C.$\dfrac{{24}}{{91}}$
D.$\dfrac{{89}}{{91}}$

Answer 1

Hint: The number of ways of selecting $r$ items from $n$ different items is determined by the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. 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. ${\text{Probability}} = \dfrac{{{\text{No}}{\text{. of favorable ways}}}}{{{\text{Total number of ways}}}}$.

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 $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
Thus the number of ways of selecting $3$ items from total $15$ items is given as:
$
\Rightarrow {\text{Total number of ways}}{ = ^{15}}{C_3} = \dfrac{{15!}}{{3!12!}} = \dfrac{{15 \times 14 \times 13 \times 12!}}{{6 \times 12!}} \\
\Rightarrow {\text{Total number of ways}} = 455{\text{ }}.....{\text{(1)}} \\
$
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:
$
\Rightarrow {\text{ No}}{\text{. of favorable ways}}{ = ^{10}}{C_3} = \dfrac{{10!}}{{3!7!}} = \dfrac{{10 \times 9 \times 8 \times 7!}}{{6 \times 7!}} \\
 \Rightarrow {\text{No}}{\text{. of favorable ways}} = 120{\text{ }}.....{\text{(2)}} \\
$
Further, we know that probability of an event is calculated by the formula:
$ \Rightarrow {\text{Probability}} = \dfrac{{{\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:
$
\Rightarrow {\text{Probability}} = \dfrac{{120}}{{455}} = \dfrac{{24 \times 5}}{{91 \times 5}} \\
\Rightarrow {\text{Probability}} = \dfrac{{24}}{{91}} \\
$

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

Note:
The formula used for combination i.e. $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ 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:
 ${ \Rightarrow ^n}{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$