Question

There are 5 duplicate and 10 original items in an automobile shop and 3 items are bought at random by a customer. The probability that none of items is duplicate, is

• A. $\frac{20}{91}$
• B. $\frac{22}{91}$
• C. $\frac{24}{91}$
• D. $\frac{89}{91}$

Answer 1

The correct option is C $\frac{24}{91}$

Total number of items = 15

Hence, if 3 items are chosen at random total number of outcomes of this selection = $n\left(S\right){=}^{15}{\text{C}}_3=\frac{15\times 14\times 13}{3!}=455$

Total number of duplicate items = 5

Total number of original items = 10

Thus, number of ways in which one can choose 3 original items from 10 = $n\left(A\right){=}^{10}{\text{C}}_3=\frac{10\times 9\times 8}{3!}=120$

Let $E$ be the event of selecting 3 original items from the automobile shop.

Thus, $P\left(E\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{120}{455}=\frac{24}{91}$

Hence, the correct answer is option C.