# Probability of forming dihydrogen with molecular weight 3 [closed]

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,

$$\begin{split} P[\ce{H2}] &= P[(\ce{^1H} \cap \ce{^2H}) \cup (\ce{^2H} \cap \ce{^1H})] \\ % &= P[\ce{^1H} \cap \ce{^2H}] + P[\ce{^2H} \cap \ce{^1H}] - % P[(\ce{^1H} \cap \ce{^2H}) \cap (\ce{^2H} \cap \ce{^1H})] \end{split} \tag{1}$$

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

$$\begin{split} P[\ce{H2}] &= P[\ce{^1H} \cap \ce{^2H}] \\ &= \gamma(\ce{^1H}) \cdot \gamma(\ce{^2H}) \\ &= 0.99972\cdot 0.00028 \\ &= 0.0002799 \end{split} \tag{4}$$

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?

• I think you may need to explain exactly to us chemists what your notation is trying to express. I wonder, though, if this is also overcomplicating it. From your other posts on SE, you obviously know plenty of maths. This is nothing more than a simple combinatorics problem: if you draw a red ball with probability 99.97% and blue ball with probability 0.03%, what are the chances that you will get a red and a blue ball if you draw 2 balls. Dec 1 '20 at 19:27
• It would be $99.97\cdot 0.03$ %, which is my answer because you'd be looking for $P[red \cap blue] = P[red | blue]\cdot P[blue] = P[red]\cdot P[blue]$ since $P[red | blue] = P[red]$. Now, if $red = ^{1}H$ and $blue = ^{2}H$, then my answer holds but not the one from my professor Dec 1 '20 at 19:33
• The order doesn't matter. Not for the hydrogen molecule, anyway. Dec 1 '20 at 19:37
• @Maurice Vicky uses set symbols describing probability. $A \cup{} B$ describes that either event $A$, or event $B$ takes place (addition of individual probabilities). $A \cap{} B$ describes if events $A$ and $B$ coincide. E.g., $A \cup{} B = \varnothing$ states both events are incompatible. You may state $A \cup{} \bar{A} = \Omega$, the event of either $A$, or the opposite of $A$ is sure to happen, probability = 1; and you may state $A \cup{} \bar{A} = \varnothing$ because coincidence of $A$ and the opposite of $A$ is zero. Maybe you are more used to $\vee$ and $\wedge$. Dec 1 '20 at 20:59
• @Vicky I understand your problem editing two concepts at once. The irritation from the chemists side is that the standard mathematical notation about the probabilities partially overlaps with the chemical notation of superscripts denoting the two most abundant isotopes of hydrogen. Plus that the first expression spans over multiple equal signs as a one-liner (which often is not helpful at all). Dec 1 '20 at 21:11

## 2 Answers

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.

You are wrong. There are only four cases (first atom is proton or deuteron, second atom is proton or deuteron), and you have to add up two cases to get all molecules with mass number of 3, i.e. the two light blue areas in the schematic diagram below (not to scale):