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A man possesses five coins, two of which are double-head (DH), one is double-tail (DT), and two are normal (N). He shuts his eyes, pick a coin at random, and tosses it. What is the probability that the lower face of the coin is a head.

He open his eyes and sees that the coin is showing heads; what is the probability that the lower face is a head? He shuts his eyes again and tosses the coin again. What is the probability that the lower face is a head? He open his eyes and sees that the coin is show heads; what is the probability that the lower face is a head?

He discards this coin, picks another at random, and tosses it. What is the probability that it shows heads?

When the man tosses a coin at the first time, the only information we have are those five coins. Let $H_1$ denote as first toss, the probability that the lower face is a head at the first toss can be $$\mathbb{P}(H_1)=\mathbb{P}(H\vert DH)+\mathbb{P}(H\vert DT)+\mathbb{P}(H\vert N)=2/5+0+1/2\times 2/5=3/5$$ At his second toss $H_2$, we know that probability of first toss is $3/5$ which will effect the second toss, so $$\mathbb{P}(H_2\vert H_1)=\frac{\mathbb{P}(H_2\cap H_1)}{\mathbb{P}(H_1)}=\frac {\mathbb{P}(DH)}{\mathbb{P}(H_1)}=(2/5)/(3/5)=2/3$$ His second toss will effect his third toss, I might get $\mathbb{P}(H_3\vert (H_2\vert H_1))$ which I don't how to calculate and don't know it is a legal expression or not. I get stuck for the remain questions.

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The law of total probability is: $$\begin{align}\mathsf P(H_1)~=~& \mathsf P(H_1\cap DH)+\mathsf P(H_1\cap DT)+\mathsf P(H_1\cap N) \\ ~=~& \mathsf P(H_1\mid DH)\mathsf P(DH)+\mathsf P(H_1\mid DT)\mathsf P(DT)+\mathsf P(H_1\mid N)\mathsf P(N) \\ =~& 1\cdot \tfrac 25 + 0\cdot\tfrac 1 5+\tfrac 1 2\cdot\tfrac 2 5 \\ =~& \frac 35\end{align}$$

So you accidentally arrived at the right value.


$\mathsf P(H_1\mid H_2) = \mathsf P(H_1\cap H_2)/\mathsf P(H_2) = \tfrac 25/\tfrac 35 = \tfrac 2 3$ as you had.


$\require{cancel}\color{red}{\xcancel{\color{black}{\mathsf P(H_3\mid(H_2\mid H_1))}}}$ is indeed not a valid construction.   There is only ever one divider between the event and the condition.

In any case, what you seek if $\mathsf P(H_3\mid H_2)$ : the probability that the lower face of the second toss is heads when given that the upper face of the first toss was heads.

Use similar reasoning as above.

$\mathsf P(H_3\mid H_2) ~=~ \dfrac{\mathsf P(H_3\cap H_2)}{\mathsf P(H_2)}~=~\dfrac{\mathsf P(H_2\cap H_3\cap DH)+0+\mathsf P(H_2\cap H_3\cap N)}{\mathsf P(H_2)}$


Now find $\mathsf P(H_3\mid H_2\cap H_4)$, the probability that the lower face of the second toss is heads given that the upper face of both tosses were each heads.

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