If I flip $1$ of $3$ modified coins $3$ times, what's the probability that I will get tails? 
We have $3$ modified coins: $M_1$ which has tails on the both sides, $M_2$ which has heads on the both sides and $M_3$ which is a fair coin. We extract a coin from the urn and we flip it $3$ times.
  
  
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*What is the probability that if I flip the coin $3$ times I will get all tails?
  
*If I got all tails at all $3$ flips what is the probability that the extracted coin is $M_3$?
  

My attempt:


*

*I have tried this way: There is a $\frac{1}{3}$ chance to get $M_1$ or $M_2$ or $M_3$. If we get $M_1$ the probability to get tails is $1$, for $M_2$ is $0$ and for $M_3$ is $\frac{1}{2}$. Then the probability to get tails at one flip is $$\frac{1}{3}\cdot 1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{2}$$ So the probability to get tails at all the $3$ flips is ${(\frac{1}{2})}^3$ which is $\frac{1}{8}$. Is this right?

*The probability seems to be intuitively $\frac{1}{3}$, but I don't know how to formally prove it.
 A: By Bayes' theorem \begin{align}P(M_3\mid TTT)=\frac{P(TTT\mid M_3)P(M_3)}{P(TTT)}=\frac{\left(\frac12\right)^3\cdot\frac13}{\frac38}=\frac19\end{align} 

Note: The denominator was calculated using the Law of total probability as is common when applying the Bayes rule. You did this in part 1. but not correctly. To see this write \begin{align}P(TTT)&=P(TTT\mid M_1)P(M_1)+P(TTT\mid M_2)P(M_2)+P(TTT\mid M_3)P(M_3)\\[0.2cm]&=1\cdot\frac13+0\cdot\frac13+\left(\frac12\right)^3\frac13\\[0.2cm]&=\frac13\left(1+\frac18\right)=\frac38\end{align}
A: Paint the double-head coin yellow on one side and red on the other.
Paint the double-tail coin blue and green.
There are 24 possible outcomes.  One of the outcomes is:  


*

*Pick double-head, toss yellow,yellow,yellow


Run through the 24 outcomes, how many of them give you three tails?
Of those outcomes, how many were with the fair coin, how many were with the double-tail coin?
A: I will extend my comment: remember that $\text{probability of A}=\frac{\text{number of A cases}}{\text{all possible cases}}$.
Then, how many ways we can get (tail, tail, tail)? If we take the fair coin with this coin we only can take (tail, tail, tail) i.e. only exist one way we can take the desired result.
But if we took the double-tail coin we take (tail, tail, tail) any time i.e. the full 8 ways that a coin can show when it is tossed three times.
And when we get the double-head coin we cant take (tail, tail, tail).
Then the total amount of ways we can take (tail, tail, tail) is just $1+8$, and the cases for the fair coin is just $1$ so the probability that you want is $1/9$.
This is a visual way to see the problem but the formal way to solve it is the answer of @JimmyR i.e. using the basic definitions and theorems of probability theory.
A: There are some correct answers here. Many use Bayes' rule, which is correct and elegant but takes getting used to. Let me try instead to help you think through this particular example, to train your intuition.
In your answer to #1 you correctly compute that the probability of one $T$ is 1/2. But that doesn't mean the probability of $TTT$ is 1/8 unless you put the coin back and choose independently again for each of the next two tosses. The way the problem is stated, you use the same coin all three times. Then the right way to compute the weighted average is
$$
\frac{1}{3}⋅1+ \frac{1}{3}⋅0+ \frac{1}{3}⋅\frac{1}{8}= \frac{3}{8}.
$$
For the second question, you know that you don't have the middle coin, so you need the probability of the first compared to the last. If you imagine that you can tell the two sides of the two-tailed coin apart, there are 8 ways to do three flips, all of which are all tails. For the fair coin, only one triple is all tails. So when you see all tails the probability that you had the all-tail coin is 8/9.
