probability
When you flip a coin four times, what is the probability that it will come up heads
exactly twice?
My calculation:
Thus the probability is: $2/ 8 = 0.25 $
but the correct answer is $0.375$. Can anyone explain why I'm wrong?
The average amount of time you need to wait til the first heads is $\frac{1}{p}$ (this is the mean of a geometric distribution with parameter $p$ - the time it takes for the first head to appear when iid coins are flipped with heads probability $p$). After you get the first heads, you wait on average another $\frac{1}{p}$ for the second heads, since the geometric distribution si memoryless. So on so forth, and you get $\frac{5}{p}$. Alternatively, you can use the negative binomial distribution as suggested in the comments.
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 $$ A \text{ xor }B = [(A\text{ is true})\lor(B\text{ is true})]\land ((A\text{ and }B)\text{ is not true}) \\= A \bigtriangleup B = \{HT, TH \} = A \cup B \setminus A \cap B = \{HH, HT, TH \} \setminus \{HH \}$$
Best Answer
Two results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is $2\times 2\times 2\times 2$ results in total. That is $2^4$ or $16$.
For the favourable case we need to count the ways to get $2$ heads and $2$ tails. The count of permutations of two pairs of symbols is: $\frac{4!}{2!2!}=6$. This is easily confirmed by just counting.
$$\Bigl|\{\mathsf {HHTT, HTHT, HTTH, THHT, THTH, TTHH}\}\Bigr|=6$$
Thus the probability is: $\tfrac{\;6}{16}$, or: $$0.375$$