Question

Ten persons with badges numbered $1$ to $10$ are in a room. If three of them are asked to leave the room. Then, the probability to have the person with the smallest badge number as $5$ among the three persons that left the room, is

• A. $\frac{3}{10}$
• B. $\frac{1}{6}$
• C. $\frac{1}{12}$
• D. $\frac{2}{5}$

Answer 1

Answer: (C)

Let $S$ be the event of selecting $3$ number from first $10$ natural number
$\therefore n(s)={}^{10}{C}_3$
Let $A$ be the event of selecting three number from $1$ to $10$ number out of which the smallest number is $5$ .
$\therefore n(A)={}^1{C}_1\times {}^5{C}_2$
$\therefore$ Required probability
$P(A)=\frac{n(A)}{n(S)}=\frac{{}^1{C}_1\times {}^5{C}_2}{{}^{10}{C}_3}$
$=\frac{1\times \frac{5\times 4}{2\times 1}}{\frac{10\times 9\times 8}{3\times 2\times 1}}=\frac{1}{12}$