Probability that first coin flip was heads given that 7 of 10 flips were heads? Let's say we flip a coin 10 times and we get heads 7 times. What is the probability that the first flip was heads? (My naïve estimation would be 7/10 but that's just intuition, trying to prove that).
 A: I would use the binomial distribution to proof it, which is not very time consuming. Let $A$ be the event, that the first flip is head. And $B$ is the event that we get 7 times head in 10 coin flips. Then it is asked for $P(A|B)=\frac{P(A\cap B)}{P(B)}$.
$P(A\cap B)$ is the probability that we get head at the first flip and 6 heads in the remaining 9 flips. The probability is $0.5\cdot \binom{9}{6}\cdot 0.5^6\cdot 0.5^3$. And $P(B)=\binom{10}{7}\cdot 0.5^7\cdot 0.5^3$. Then the fraction is
$$P(A|B)=\frac{0.5\cdot \binom{9}{6}\cdot 0.5^6\cdot 0.5^3}{\ \ \  \quad \binom{10}{7}\cdot 0.5^7\cdot 0.5^3}=\frac{ \binom{9}{6}\cdot \require{cancel} \cancel{0.5^7}\cdot \require{cancel} \bcancel{0.5^3}}{\binom{10}{7}\cdot \require{cancel} \cancel{0.5^7}\cdot \require{cancel} \bcancel{0.5^3}}=\frac{ \binom{9}{6}}{\binom{10}{7}}$$
Using $\binom{n}{k}=\binom{n}{n-k}$ and $\binom{m}{r}=\frac{m\cdot (m-1)\cdot (m-2)\cdot \ldots \cdot (m-r+1)}{1\cdot 2\ \cdot \ldots \cdot r} (*)$ we obtain
$$\frac{\binom{9}{6}}{\binom{10}{7}}=\frac{ \binom{9}{3}}{\binom{10}{3}}=\frac{9\cdot 8\cdot 7}{1\cdot 2\cdot 3}\cdot \frac{1\cdot 2\cdot 3}{10\cdot 9\cdot 8}=\dots$$
Some steps you can omit, if you wish.
(*) The key point at this formula is that the number of factors are $r$ for both, the numerator and the denominator. At the numerator it starts with $m$ and then decreases. At the denominator it starts with $1$ and then increases.
