Assuming every question is equally likely there are nCr(50,3) = 50\*49\*48/(3\*2) = 19'600 possibilities.
Assuming, one of the choices has to be the question, there are nCr(49,2)\*1= 49\*48/2= 1176 possibilities.
Now, these two events are assumed to be independent ie the questions of the first exam don't influence the questions of the second exam.
This means, the probability of this event happening twice in a row is equal to the probability of it happening once squared.
The probability of this happening is
>P(x) = 1176/19600 => P(x)\^2= (1176/19600)\^2=0.0036 = 0.36%