Probability of forming dihydrogen with molecular weight 3
Extending / reformulating Karsten Theis' answer: A probabilistic approach is to draw a tree about drawing two spheres where event $A$ (normal hydrogen, $\ce{^1H}$) dominates, but event $B$ (the next abundant isotope, $\ce{^2H}$) complements, $P(\ce{^1H} \cup{} \ce{^2H}) = 1 = \Omega{}$ and eventually to multiply the probabilities for each path (e.g., $P(\ce{^1H^2H})$ as shown on the left hand side).

Yet because in your case both paths $P(\ce{^1H^2H})$ and $P(\ce{^2H^1H})$ yield a dihydrogen molecule of mass 3, these individual probabilities need to be summed up.