A three digit number, which is a multiple of 11 is chosen at random. Probability that the number so chosen is also a multiple of 9, is equal to

A= Three digit numbers multiple of 11 are 

A={110, 121,...,990}

n(A) = 81.
n= number also divisible by 9 

$A\cap B=$Three digit numbers are divisible by 99,
$A\cap B=${ 198, 297, ..., 990} 

$n(A\cap B)=9$
So, required probability $P(B/A)=\frac{n(A\cap B)}{n\left(A\right)}=\frac{9}{81}=\frac{1}{9}$