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A person has five coins, two of which are double headed, one of which is double-tailed, and the remaining two are normal. The person shuts their eyes, picks a coin at random and tosses it

Suppose now the outcome of the first toss was heads, we discard the coin and pick a new one (blindly) from the remaining coins and toss it. What is the (conditional) probability the outcome is heads?

I tried to compute P(H2|H1) when the first coin is double headed + P(H2|H1) when the first coin is normal But it didn't work out (I got 13/16)

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    $\begingroup$ What did you try? Add it to the question, even if it's wrong. The idea is to help you solve the problem, but not solve it for you. $\endgroup$ – rtybase May 12 at 10:30
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    $\begingroup$ @rtybase see edit! $\endgroup$ – Katie May 12 at 10:37
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Given that we got heads on the first toss, there is a $\frac23$ probability of having chosen one of the double headed coins. Removing one of the double headed coins, there is a $\frac12$ probability of getting heads on the second toss.

Given that we got heads on the first toss, there is a $\frac13$ probability of having chosen one of the regular coins. Removing one of the regular coins, there is a $\frac58$ probability of getting heads on the second toss.

Given that we got heads on the first toss, there is a $\frac23\cdot\frac12+\frac13\cdot\frac58=\frac{13}{24}$ probability of getting heads on the second toss.

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A very elaborate version, but hopefully with enough details.

Let's model the coins, say $K_1=(H|H)$ (double head), $K_2=(H|H)$, $K_3=(T|T)$, $K_4=(T|H)$ and $K_5=(T|H)$. Now $$P(H_2|H_1)=\frac{P(H_2 \cap H_1)}{P(H_1)}$$ Probability of getting heads in round $1$ is $P(H_1)=\frac{6}{10}=\color{red}{\frac{3}{5}}$, i.e. there are $6$ of $H$ out of $10$ possible $\{H,T\}$ in the coin model above.


As a side note, this can also be calculated as $$P(H_1)= P(\text{pick } K_1)\cdot P(K_1=H)+\\ P(\text{pick } K_2)\cdot P(K_2=H)+\\ P(\text{pick } K_3)\cdot P(K_3=H)+\\ P(\text{pick } K_4)\cdot P(K_4=H)+\\ P(\text{pick } K_5)\cdot P(K_5=H)=\\ \frac{1}{5}\cdot 1 + \frac{1}{5}\cdot 1 + \frac{1}{5}\cdot 0+\frac{1}{5}\cdot \frac{1}{2}+\frac{1}{5}\cdot \frac{1}{2}=\color{red}{\frac{3}{5}}$$


With regards to (to shorten the text, $K_i=H$ means "pick coin $K_i$ and tossing result is $H$") $$P(H_2 \cap H_1)=P\left(H_2 \cap ((K_1=H)\cup (K_2=H)\cup (K_4=H)\cup (K_5=H))\right)$$ (obviously $K_3=H$ is empty) and $$P(H_2 \cap H_1)=P( (H_2 \cap (K_1=H)) \cup \\ (H_2 \cap K_2=H)) \cup \\ (H_2 \cap (K_4=H)) \cup \\ (H_2 \cap K_5=H))))=\\ P(H_2 \cap (K_1=H))+\\ P(H_2 \cap (K_2=H))+\\ P(H_2 \cap (K_4=H))+\\ P(H_2 \cap (K_5=H))=...$$ given we remove the coin after the $1$st toss (thus we are left with $8$ of possible $\{H,T\}$) $$..=\frac{4}{8}\cdot\frac{1}{5}\cdot1+\frac{4}{8}\cdot\frac{1}{5}\cdot1+ \frac{5}{8}\cdot\frac{1}{5}\cdot\frac{1}{2}+ \frac{5}{8}\cdot\frac{1}{5}\cdot\frac{1}{2}=\frac{13}{40}$$ and finally $$P(H_2|H_1)=\frac{\frac{13}{40}}{\frac{3}{5}}=\frac{13}{24}$$

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I tried to compute P(H2|H1) when the first coin is double headed + P(H2|H1) when the first coin is normal . But it didn't work out (I got 13/16)

You did not account for the probabilities that you might have chosen those coins.

Let $D_H,F$ be the event that the first coin chosen was Double-Head or Fair, respectively.

$$\begin{align}\mathsf P(H_2\mid H_1)&=~\mathsf P(H_2\mid D_H)~\mathsf P(D_H\mid H_1)+\mathsf P(H_2\mid F)~\mathsf P(F\mid H_1)\end{align}$$

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