Calculating probability of pressing button three times and hearing sound at the third press Consider the situation: 
There are two mysterious buttons in front of you. One of the buttons is harmless, whenever you press it, nothing happens. The other button is mostly harmless, when you press it,nothing happens with probability 2/3 but with probability 1/3, pressing the button results in a loud siren! Unfortunately you don't know which button is which.
I have to answer the following: 
Suppose you pick a button at random and press it twice. Suppose nothing happens either
time you press it. What's the probability when you press it a third time that you hear the siren?
I am stuck trying to calculate this. As of right now all I have calculated is the event that you pick a button and press it twice and do not hear a siren (call this event S). 
So P(S) = (1/2) + (1/2)((2/3)^2) = 13/18
(I used the law of total probability to calculate the above value)
Now I want to calculate the event that I hear a siren on the third try after not hearing it the two tries before. How would I go about doing so? I can't seem to wrap my head around this problem. 
 A: Let's say button $A$ is harmless, but button $B$ has a $1/3$ chance of yelling at you. If you press $A$ twice, there's a $1/1$ chance nothing happens. If you press $B$ twice, there's a $4/9$ chance that nothing happens.
Now if you press a random button twice and nothing happens, then the probability of you having chose $A$ is
$$\frac{1}{1+(4/9)}=9/13$$
Probability of having chose $B$ is
$$\frac{4/9}{1+(4/9)}=4/13$$
EXPLANATION:

Here are the probabilities for $A$ and $B$ for not hearing a siren after $2$ presses. Don't really know which button you chose, but you know you are in this blue area. The probability that you are in $A$'s blue area is:
$$\frac{1}{1+(4/9)}=9/13$$
Now if you chose $A$, the probability of hearing a siren on the $3$rd try is $0$. If you chose $B$, then the probability of hearing the siren is $1/3$.
The probability of hearing the siren on the $3$rd press after choosing a random button and not hearing the siren on the first two presses, is
$$\frac9{13}\cdot0+\frac4{13}\cdot\frac13=\frac4{39}$$
A: Let $T$ be the event that you hear a siren on the third try (but not the first two tries). Then:
\begin{align*}
\text{Pr}(T \mid S) &= \frac{\text{Pr}(S \cap T)}{\text{Pr}(S)} \\
&= \frac{\text{Pr}(T)}{\text{Pr}(S)} \\
&= \frac{\frac{1}{2} \cdot (\frac{2}{3})^2(\frac{1}{3})}{\frac{1}{2} + \frac{1}{2} \cdot (\frac{2}{3})^2} \\
&= \frac{4}{39}
\end{align*}
