Question

Three numbers are chosen at random without replacement the probability that the minimum of the chosen numbers is $3$ or their maximum is $7$ is

• A. $\frac{11}{20}$
• B. $\frac{1}{40}$
• C. $\frac{13}{40}$
• D. $\frac{1}{15}$

Answer 1

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

Number of elements in sample space ${=}^{10}{C}_3=120$

According to the question $3$ & $7$ must get chosen and the third no. should be between them (i.e $4$ or $5$ or $6$) as per condition .

So, no of favorable cases ${}^3{C}_1=3$

Therefore required probability $=\frac{3}{120}=\frac{1}{40}$

Hence, the option $B$ is the correct answer.