probability
If a coin has a $0.7$ chance of landing tails. What is the probability of getting EXACTLY $5$ heads in $10$ flips?
I know that the probability is $\frac{63}{256}$ if the coin is fair but I cannot work out how to do this problem. This is all i have so far:
$$
\frac{(10!)}{2^{10}\cdot 5!\cdot5!} = \frac{252}{1024} = \frac{63}{256}
$$
Your question is related to the binomial distribution.
You do $n = 10$ trials. The probability of one successful trial is $p = \frac{1}{2}$. You want $k = 3$ successes and $n - k = 7$ failures. The probability is:
$$ \binom{n}{k} p^k (1-p)^{n-k} = \binom{10}{3} \cdot \left(\dfrac{1}{2}\right)^{3} \cdot \left(\dfrac{1}{2}\right)^{7} = \dfrac{15}{128} $$
One way to understand this formula: You want $k$ successes (probability: $p^k$) and $n-k$ failures (probability: $(1-p)^{n-k}$). The successes can occur anywhere in the trials, and there are $\binom{n}{k}$ to arrange $k$ successes in $n$ trials.
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
The number of ways you can arrange five heads and five tails (first toss to last) is $_{10}C_{5}$.
The probability of getting five of each, in one particular order (say all tails, then all heads), is $0.3^50.7^5$.
Can you take it from here?