binomial distributionprobability
A coin is weighted to 1/12 heads and 11/12 tails.
What is the probability you get exactly 2 heads in 5 coin flips?
I think it's:
(Probability of 2 heads) * (Probability of 3 tails) * (number of ways this can occur)
= (1/12 * 1/12) * (11/12 * 11/12 * 11/12) * (5!/(3!*2!))
= 0.05348990484
The symmetric difference, of exclusive or, is the probability of getting exactly one head. What you were calculating was the union, which is the probability of at least one head.
$\newcommand{\P}{\mathbb{P}}$In mathematics, "or" means "either $A$ is true, $B$ is true, or both $A$ and $B$ are true. See this Wikipedia page which has nice diagrams: https://en.wikipedia.org/wiki/Logical_disjunction.
In other words $\mathbb{P}(A\text{ or }B)= \mathbb{P}(A\text{ only})+ \P(B\text{ only})+ \P(\text{both }A\text{ and }B)$.
If I am interpreting your question correctly, $A = \{HH, HT \}$ and $B = \{ HH, TH\}$.
This means that $$A\text{ or }B = (A \text{ is true}) \lor (B \text{ is true}) = A \cup B = \{HH, HT, TH \}$$
Therefore $\P(A \text{ or }B) = \frac{3}{4}$ and the formula which you claim does not work does work. $$\P(A)=\frac{1}{2}, \quad \P(B)=\frac{1}{2}, \quad P(A \cap B)=\frac{1}{4}, \quad \frac{1}{2}+\frac{1}{2} - \frac{1}{4} = \frac{3}{4}$$
The word "or" is often used in English to mean "exclusive or" or "XOR", which means "either $A$ only, or $B$ only, but not both". This corresponds to the set operation of symmetric difference, rather than union. However, in probability, it is always union, and not symmetric difference, that is meant when the term "or" is used. This makes most mathematical expressions and statements much easier to state and formulate.
Here is a picture below of symmetric difference; contrast it with that of union you see above:
This means that $$ A \text{ xor }B = [(A\text{ is true})\lor(B\text{ is true})]\land ((A\text{ and }B)\text{ is not true}) \\= A \bigtriangleup B = \{HT, TH \} = A \cup B \setminus A \cap B = \{HH, HT, TH \} \setminus \{HH \}$$
You need to count the arrangements for exactly 3 heads and 3 tails.
That is the permutations of $\sf HHHTTT$.
As you state in subsequent comments, that is $6!/3!^2$.
Best Answer
As has been stated in the comments, your calculation is correct.