A five digit number is chosen at random. The probability that all the digit are distinct and digits at odd places are odd and digits at even place are even, is

Total number of 5-digit numbers = 9 x 10x 10x10x 10 = 90000

n(S)= 90,000
E = odd digits in odd places and even digits in even places 

E = The 5 odd digits {1,3,5,7,9} can be arranged in the three odd places in 5x4x3 ways

The 5 even numbers {0,2,4,6,8} can be arranged in 2 even places in 5x4 ways

n(E) = 5x5x4x4x3=1200
Thus. required probability $P\left(E\right)=\frac{1200}{90000}=\frac{1}{75}$