Three numbers are chosen at random without replacement from ${1,2,....10}$ . The probability that the minimum of the chosen numbers 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,
$P(A)=P$(choosing 3 and two other numbers from 4 to 10)
$=\cfrac { ^{ 7 }{ { C }_{ 2 } } }{ ^{ 10 }{ { C }_{ 3 } } } =\cfrac { 7\times 6 }{ 2 } \times \cfrac { 3\times 2 }{ 10\times 9\times 8 } =\cfrac { 7 }{ 40 }$
$P(B)=P$(choosing 7 and two other numbers from 1 to 6)
$=\cfrac { ^{ 6 }{ { C }_{ 2 } } }{ ^{ 10 }{ { C }_{ 3 } } } =\cfrac { 6\times 5 }{ 2 } \times \cfrac { 3\times 2 }{ 10\times 9\times 8 } =\cfrac { 1 }{ 8 }$
$P\left( A\cap B \right) =P$(Choosing 3 and 7 and one other number from 4 to 6)
$=\cfrac { 3 }{ ^{ 10 }{ { C }_{ 3 } } } =\cfrac { 3\times 3\times 2 }{ 10\times 9\times 8 } =\cfrac { 1 }{ 40 }$
Now $P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) \\ =\cfrac { 7 }{ 40 } +\cfrac { 1 }{ 8 } -\cfrac { 1 }{ 40 } =\cfrac { 11 }{ 40 }$