Question
There are 10 boxes in front of me, the $k^{th}$ box contains coins that lands heads with probability $\frac{1}{k}$ when flipped.
I pick a random box, then a coin from the box and flip it. The coin lands heads. What is the probability that it was from the first box?
I do it again, and it lands heads again. What is the probability that it was from the first box?
I do it again, and it lands heads again. What is the probability now that it is from the first box?
My Attempt At a Solution
Let $F$ be the event that the coin is from the first box. $P(F)=\frac{1}{10}$.
Let $K$ be the event that the coin is from the $k^{th}$ box. $P(K)=\frac{1}{10}$.
Let $N$ be the event that the coin is from the first box. $P(F)=\frac{4}{5}$.
Using the definition of conditional probability: $$P(A|B)=\frac{P(A \cap B)}{P(B)} $$
$$P(F|H)=\frac{P(F \cap H)}{P(H)} =\begin{cases} \frac{\frac{1}{10}\frac{1}{2}}{\frac{1}{2}}=\frac{1}{10} \quad \text{ for a fair coin} \\ \frac{\frac{1}{10}\frac{1}{1}}{\frac{1}{1}}=\frac{1}{10} \quad \text{ for an unfair coin with probability } \frac{1}{k}; k=1\end{cases} $$
$$P(F|(H\cap H))=\frac{P(F \cap (H\cap H))}{P(H\cap H)} \begin{cases} \frac{\frac{1}{10}\frac{1}{4}}{\frac{1}{4}}=\frac{1}{10} \quad \text{ for a fair coin} \\ \frac{\frac{1}{10}\frac{1}{1}^2}{\frac{1}{1}^2}=\frac{1}{10} \quad \text{ for an unfair coin with probability } \frac{1}{k}; k=1\end{cases} $$
$$P(F|(H\cap H\cap H))=\frac{P(F \cap (H\cap H\cap H))}{P(H\cap H\cap H)} \begin{cases} \frac{\frac{1}{10}\frac{1}{8}}{\frac{1}{8}}=\frac{1}{10} \quad \text{ for a fair coin} \\ \frac{\frac{1}{10}\frac{1}{1}^3}{\frac{1}{1}^3}=\frac{1}{10} \quad \text{ for an unfair coin with probability } \frac{1}{k}; k=1\end{cases} $$
So the probability is always $\frac{1}{5}$?
Thank you for any help/hints and your time.
You have calculated the denominator incorrectly. We must divide by the total probability that heads was obtained from any of the boxes that could have been selected.
Let's define our variables as follows: Let $B_k$ be the event that the ball is from the $k$th box, $1 \leq k \leq 10$. Let $H$ be the event that the coin lands heads.
Since the box from which the coin is extracted is chosen at random, $$\Pr(B_k) = \frac{1}{10}, 1 \leq k \leq 10$$ Since the probability that a coin from the $k$th box lands heads is $1/k$, $$\Pr(H \mid B_k) = \frac{1}{k}, 1 \leq k \leq 10$$
The probability that the coin is from the first box given that it lands heads is $$Pr(B_1 \mid H) = \frac{Pr(H \cap B_1)}{\Pr(H)}$$ where $\Pr(H)$ is the total probability that heads was obtained from any of the boxes that could have been selected. Therefore,
\begin{align*} Pr(B_1 \mid H) & = \frac{\Pr(H \cap B_1)}{\Pr(H)}\\ & = \frac{\Pr(B_1)\Pr(H \mid B_1)}{\sum_{k = 1}^{10} \Pr(B_k)\Pr(H \mid B_k)}\\ & = \frac{\frac{1}{10} \cdot 1}{\sum_{k = 1}^{10} \frac{1}{10}\frac{1}{k}}\\ & = \frac{\frac{1}{10}}{\frac{1}{10} \sum_{k = 1}^{10} \frac{1}{k}}\\ & = \frac{1}{\sum_{k = 1}^{10} \frac{1}{k}} \end{align*}
The probability that the coin is from the first box given that it lands heads on the first two throws is \begin{align*} \Pr(B_1 \mid H \cap H) & = \frac{\Pr(B_1 \cap H \cap H)}{\Pr(H \cap H)}\\ & = \frac{\Pr(B_1)\Pr(H \mid B_1)\Pr(H \mid H \cap B_1)}{\sum_{k = 1}^{10} \Pr(B_k)\Pr(H \mid B_k)Pr(H \mid H \cap B_k)}\\ & = \frac{\frac{1}{10} \cdot 1 \cdot 1}{\sum_{k = 1}^{10} \frac{1}{10} \cdot \frac{1}{k} \cdot \frac{1}{k}}\\ & = \frac{\frac{1}{10}}{\frac{1}{10}\sum_{k = 1}^{10} \frac{1}{k^2}}\\ & = \frac{1}{\sum_{k = 1}^{10} \frac{1}{k^2}} \end{align*} I leave the calculations and the third part of the problem to you.