You are given a coin. The coin is unevenly manufactured, such that side 1 is heavier than side 2. We denote the weight of side 1 by $x$ and of side 2 by $y$. You throw the coin $100$ times and $30$ times side 2 faces downwards. Compute the probability that $Pr(y \leq 2x| \text{side 2 30 times down})$.
Ok, so the probability that side 1 faces downwards is $Pr(\text{side 1 down})=\frac{x}{x+y}$ and $Pr(\text{ side 2 down }) = \frac{y}{x+y}$. The probability that if you throw $100$ times side 2 is downwards is $Pr(\text{ side 2 $30$-times down})=\binom{100}{30}(\frac{y}{x+y})^{30}(\frac{x}{x+y})^{70}$. Now we can obtain something like $Pr(\text{side 2 30-times down}|y\leq 2x)\leq \binom{100}{30}2^{30}\frac{x^{100}}{(x+y)^{100}}$. I tried playing around with different formulations of conditioned probability like $Pr(A|B)=\frac{Pr(A)}{Pr(B)}Pr(B|A)$, but since I do not know $P(y\leq 2x| \text{side 2 30-times dowm})$ I am stuck. Does someone have an idea?
$P(y\leq 2x\mid \text{side 2 30-times down})$ is what you're really asked to calculate in the exercise, the way I read it. You are supposed to take the results of the flipping experiment into account when you calculate the probability they ask for. So it's entirely expected that you don't know it, and that's fine.
The big issue here, however, is that you don't know the prior, entirely unconditional, $P(y\leq 2x)$. Without that, you can't get anywhere.
The skill and intention of the coin minter will change the initial probability, and thus the conditional probability. If one coin in a million has $y\leq2x$, then $P(y\leq 2x\mid \text{side 2 30-times down})$ is still pretty small. If the minter tries to achieve $y=2x$ on purpose, or is just very sloppy in their work, then $P(y\leq 2x\mid \text{side 2 30-times down})$ is significantly bigger.