Two friends A and B have equal number of sons. There are 3 cinema tickets which are to be distributed among the sons of A and B. The probability that all the tickets go to the sons of B is 1/20. the no. of sons each of them having is

Suppose that each of A, B have $x$ number of sons. 

The number of ways of distributing 3 tickets among all of them is $2x_{C_3}$ and the number of ways of distributing 3 tickets among sons of B is $x_{C_3}$

Hence, the probability of distributing all tickets to the sons of B is $\frac{x_{C_3}}{2x_{C_3}}$

Given $\frac{x_{C_3}}{2x_{C_3}}=\frac{1}{20}$

Simplify

$\begin{array}{rcl}\frac{x_{C_3}}{2x_{C_3}}& =& \frac{1}{20}\\ \frac{{\displaystyle \frac{x\left(x-1\right)\left(x-2\right)}{6}}}{{\displaystyle \frac{2x\left(2x-1\right)\left(2x-2\right)}{6}}}& =& \frac{1}{20}\end{array}$

cancel out the common terms $x\left(x-1\right)$ in both numerator and denominator

we get 

$\begin{array}{rcl}\frac{x-2}{4\left(2x-1\right)}& =& \frac{1}{20}\\ 5x-10& =& 2x-1\\ 3x& =& 9\\ x& =& 3\end{array}$

Therefore, the number of childrent to A is $\boxed{3}$