Answer is (C)
There are $12$ houses on one side of a street are numbered with even numbers.
In which $5$ houses are strictly greater than Number $14.$
And remaining $7$ houses are numbered smaller than $14$ (i.e. including $14$)
No of way of choosing at least $2$ of these newspapers are delivered to houses with numbers strictly greater than $14.$
$^{5}C_{3} +^{5}C_{2}\times ^{7}C_{1} =80$
Total way of choosing $3$ houses$=^{12}C_{3} =220$
So Required probability=$\dfrac{80}{220}=\dfrac{4}{11}.$