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
$P(A)=\frac{n(A)}{n(S)}=\frac{{ }^{1} C_{1} \times{ }^{5} C_{2}}{{ }^{10} C_{3}}$
$=\frac{1 \times \frac{5 \times 4}{2 \times 1}}{\frac{10 \times 9 \times 8}{3 \times 2 \times 1}}=\frac{1}{12}$