Question 385297

Total number of 3-digit numbers that can be formed = 6P3 = 120


now we find no. of 3-digit numbers that is even and greater than 900 .......


number should be in this form  :   9XE

 where E is an even number and X is any digit except 9 and digit on E.


 no. of ways to put an even digit at E i.e third place = 3  (4,6 or 8)

no. of ways to put any digit at X i.e second place= 4  
                               (except 9 and which at E )

no. of ways to put 9 at first place = 1



total no. = 1* 3 * 4 = 12 
 



thus, required probability =  12/120 = 1/10 

   = 0.1