Question

Four distinct integers are picked at random from $\{0,1,2,3,4,5,6\}.$ If the probability that among those selected, the second smallest is $3,$ is $p,$ then $p$ is equal to

• A. $\frac{9}{35}$
• B. $\frac{1}{35}$
• C. $\frac{3}{35}$
• D. $\frac{6}{35}$

Answer 1

The correct option is A $\frac{9}{35}$

Given that the second smallest is $3.$
So, select any one from $\{0,1,2\}$ by ${}^3{C}_1$ ways.
Now, select any two numbers from $\{4,5,6\}$ by ${}^3{C}_2$ ways.
Favourable cases $={}^3{C}_1\cdot {}^3{C}_2$
Total cases $={}^7{C}_4$
Required probability, $p=\frac{{}^3{C}_1\cdot {}^3{C}_2}{{}^7{C}_4}=\frac{9}{35}$