Solution:
Solution
Required probability $$=$$ probability of drawing 1 White or 2 or 3 or 4 or 5 or 6 White $$=(\dfrac { 1 }{ 6 } *\dfrac { _{ 1 }^{ 6 }{ C } }{ _{ 1 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 2 }^{ 6 }{ C } }{ _{ 2 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 3 }^{ 6 }{ C } }{ _{ 3 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 4 }^{ 6 }{ C } }{ _{ 4 }^{ 10 }{ C } } )+(\dfrac { 1 }{ 6 } *\dfrac { _{ 5 }^{ 6 }{ C } }{ _{ 5 }^{ 10 }{ C } } )=\dfrac { 251 }{ 1260 } $$
This is closest to $$\dfrac { 1 }{ 5 } $$
