Given set S={1,2,3,...8} Choosing 3 numbers from 8 numbers can be done 8C3 ways. Choosing 3 numbers from 8 numbers while minimum no is 3 can be done 1×5C2 ways. ∴ Probability P(min=3)=
1x5C2
8C3
Choosing 3 numbers from 8 numbers while maximum no is 6 can be done 1×5C2 ways. ∴ Probability P(max=6)=
1×5C2
8C3
Choosing 3 numbers from 8 numbers while minimum number 3 and maximum no is 6 can be done 1×2C1×1 ways. ∴P(min=3∩max=6)=
1×2C1×1
8C3
The probability that their minimum is 3 , given that their maximum is 6 , is : P(