From the set of all positive integral divisors of  $(2^3.3^4{.5}^2{.7}^1)$ one number ‘x’ is selected at random. If ‘x’ is even, then the probability that it is a multiple of 10 is 

Let $n=(2^3\times 3^4\times 5^2\times 7^1)$
Number of even positive integral divisors of $n=(3\times 5\times 3\times 2)=a$  
Number of positive integral divisors of ‘n’ where each divisor is a multiple of 
10 = [number of all positive integral divisors of  $\frac{n}{10}=2^2{.3}^4{.5}^1{.7}^1$]

$=(3\times 5\times 2\times 2)=b$
$\therefore$  required probability = (b/a) = 2/3