Question

Three numbers are chosen at random from the set $1,2,3,4,5,6,7,8,9,10$. The probability that the smallest of the three numbers chosen is seven is

• A. $\frac{1}{40}$
• B. $\frac{1}{10}$
• C. $\frac{1}{8}$
• D. $\frac{7}{8}$

Answer 1

The correct option is A $\frac{1}{40}$

Since the smallest of the $3$ numbers is $7$, it implies that I need to find two numbers larger than $7$.
This can be done in ${}^3{C}_2=3$ ways.
Hence, required probability = $\frac{3}{{}^{10}{C}_3}=\frac{3}{120}=\frac{1}{40}$