Question

Three numbers are chosen at random without replacement from the set $A=\{x|1\le x\le 10,xϵN\}$. The probability that the minimum of the chosen numbers is 3 and maximum is 7, is

• A. $\frac{1}{12}$
• B. $\frac{1}{15}$
• C. $\frac{1}{40}$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{40}$

The minimum number has to be 3 and maximum 7
Let the event of such kind be denoted by E
then E is the combination of 3,7 and any one of {4,5,6}
Also note that replacement is not allowed
$\Rightarrow P\left(E\right)=(3!)P\left(3\right)\times P\left(7\right)\times \frac{{}^3{C}_1}{10}$
$\Rightarrow P\left(E\right)=6\times \frac{1}{10}\times \frac{1}{9}\times \frac{3}{8}=\frac{1}{40}$