AIEEE 2012 Solved Paper

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(‌

min=3

max=6

)
=‌

P(min=3∩max=6)

P(max=6)

=‌

‌ ‌2C1 ‌8C3

‌5C2 ‌8C3

=‌

‌2C1

‌5C2

=‌

1

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