Probability of forming dihydrogen with molecular weight 3 During a lecture, my professor told us that the probability $P$ to form a hydrogen molecule $\ce{H2}$ with mass number 3 could be calculated out of the abundances, $\gamma$, of the isotopes of this element: $\ce{^1H}$ (normal hydrogen), $\ce{^2H}$ (deuterium), etc. So,
$$
\begin{split}
P[\ce{H2}] &amp;= P[(\ce{^1H} \cap \ce{^2H}) \cup (\ce{^2H} \cap \ce{^1H})] \\
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           &amp;= P[\ce{^1H} \cap \ce{^2H}] + P[\ce{^2H} \cap \ce{^1H}] - 
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              P[(\ce{^1H} \cap \ce{^2H}) \cap (\ce{^2H} \cap \ce{^1H})]
\end{split}
\tag{1}$$
But since
$$(\ce{^1H} \cap \ce{^2H}) \cap\ (\ce{^2H} \cap \ce{^1H}) = \ce{^1H} \cap \ce{^2H}\tag{2}$$
and
$$P[\ce{^1H} \cap \ce{^2H}] + P[\ce{^2H} \cap \ce{^1H}] = 2P[\ce{^1H} \cap \ce{^2H}]\tag{3}$$
we end up with
$$
\begin{split}
P[\ce{H2}] &amp;= P[\ce{^1H} \cap \ce{^2H}] \\
       &amp;= \gamma(\ce{^1H}) \cdot \gamma(\ce{^2H}) \\
       &amp;= 0.99972\cdot 0.00028 \\
       &amp;= 0.0002799
\end{split}
\tag{4}$$
This is my result while for my professor, it should be twice that value. It's like he's forgetting the term
$$
P[(\ce{^1H} \cap \ce{^2H})\ \cap\ (\ce{^2H} \cap \ce{^1H})]
$$
Am I right?
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.