Three numbers are chosen at random without replacement from $\left\{1,2,3,...,10\right\}$ The probability that minimum of the chosen number is $3$ or their maximum is $7,$ is

Let $A$ and $B$ denote the following events: 

      $A :$ minimum of the chosen number is $3$

      $B :$maximum of the chosen number is $7$

We have

             $P\left(A\right)=P$ (choosing $3$ and two other numbers from $4$ to $10$) 

                     $=\frac{{ }^7{C}_2}{{ }^{10}{C}_3}=\frac{7\times 6}{2}\times \frac{3\times 2}{10\times 9\times 8}=\frac{7}{40}$

            $P\left(B\right)=P$ (choosing $7$ and two other numbers from $1$ to $6$) 

                    $=\frac{{ }^6{C}_2}{{ }^{10}{C}_3}=\frac{6\times 5}{2}\times \frac{3\times 2}{10\times 9\times 8}=\frac{1}{8}$  

   $P(A\cap B)=P$( choosing $3$ and $7$ and one other number from $4$ to $6$).

                   $=\frac{3}{{ }^{10}{C}_3}=\frac{3\times 3\times 2}{10\times 9\times 8}=\frac{1}{40}.$

Now,
             $\begin{array}{l}P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\\ =\frac{7}{40}+\frac{1}{8}-\frac{1}{40}=\frac{11}{40}\end{array}$