Let the number of children in each family be $n$.

Then:

• Family $A$ has $n$ children
• Family $B$ has $n$ children
• Total children $= 2n$

Total ways to distribute 3 tickets

Since no child gets more than one ticket, distributing 3 identical tickets is equivalent to choosing 3 children out of $2n$.

So, total number of ways: $\binom{2n}{3}$

Favourable ways: all 3 tickets go to family $B$

We must choose all 3 children from the $n$ children of family $B$.

Number of favourable ways: $\binom{n}{3}$

Form the probability equation

Given probability: $\frac{\binom{n}{3}}{\binom{2n}{3}} = \frac{1}{12}$

Expand the combinations

$\frac{\frac{n(n-1)(n-2)}{6}}{\frac{2n(2n-1)(2n-2)}{6}} = \frac{1}{12}$

Cancel $6$: $\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} = \frac{1}{12}$

Simplify

Note that: $2n-2 = 2(n-1)$

So denominator becomes: $2n(2n-1)\cdot 2(n-1) = 4n(n-1)(2n-1)$

Hence, $\frac{n(n-1)(n-2)}{4n(n-1)(2n-1)} = \frac{1}{12}$

Cancel $n(n-1)$: $\frac{n-2}{4(2n-1)} = \frac{1}{12}$

Solve for $n$

Cross-multiplying: $12(n-2) = 4(2n-1)$

$12n - 24 = 8n - 4$

$4n = 20$

$n = 5$

Check with options

The number of children in each family is: $\boxed{5}$

This corresponds to Option C.