Probability... pls explain?

The last 3 digits are 1 and none of the digits is 0

Number is in this form -- -- -- -- 1 1 1.

we will find the probablity of having no prime digit and one prime digit

The set of numbers to choose { 2,3,4,5,6,7,8,9}

Total no. of possible numbers = 8^4 ( Each position of the number can be one of the possible 8 numbers.

Probablity of having no prime digit.

The set of numbers to choose {4,6,8,9}

Total no of numbers formed = 4^4


Probablity of having one prime digit.

The set of numbers to choose {2,3,5,7}

we have to select one from this set and 3 from non prime set

You can select one number from 4 in 4C1 ways = 4

The number being selected can take any of the four places availabe

The reamining 3 places can be filled in 4^3 ways

Total numbers formed that way = 4*4*4^3 = 4*4^4

1- ( 4^4 + 4*4^4)/8^4

= 11/16

Thanks
raama