Average time ant needs to get out to the woods An ant has three passages to choose from: 


*

*Passage A takes 7 minutes to get ant out of the ant house to the woods.

*Passage B takes 8 minutes to get ant back to the starting point where he is.

*Passage C takes 12  minutes to get ant back to the starting point where he is.
The ant chooses a passage randomly until it gets out of the ant house to the woods.
How to calculate the expected average time ant needs to get out?
Does simple mean value (7 + 8 + 12) / 3 = 9 answer the question? 
 A: Hello donkordr and welcome to CV. 
No, the mean does not answer the question unfortunately.
You can read your problem as a Markov Chain with two states: woods, and ant house. 
Now, your transition probabilities are: 


*

*From the woods: (does not really matter as it's the goal) $p=1$ to stay in the woods

*From the ant house: $p_1=\frac{2}{3}$ to stay at the ant house and $p_2=\frac{1}{3}$ to go to the woods


We want to know how many movements it takes on average to reach the woods - and you can do so as explained here 
This results in an average of 3 movements. 
Of these 3 movements, one and only one will be the one leading to the woods, while the rest will be any of the others. The average time of a movement that does not go to the woods is $(8+12)/2 = 10$ minutes. 
As a result, the average time before you reach the woods is $27$ minutes: $7$ from the last step, and $2\cdot10$ from the previous ones. 
PS - there are other ways to make this computation with intermediate states that might be cleaner, but this seemed easy enough.
A: Let me join with an explanation which, to me at least, seems even easier:
First, observe that the paths $B$ and $C$ can be joined into a single path $X$, with the passage time equal to the mean times of $B$ and $C$, i.e. $10$ minutes. This is due to the fact that these paths have the same probability. If they didn't, we'd need to take a weighted average. The probability of taking the path $X$ is the sum of probabilities of taking either the path $B$ or $C$: $P(X) = P(B) + P(C) = 2/3$.
Now, there are the following ways of getting to the woods:
$$\begin{array}{lrr}
i & \text{path}_i & P(i) & T(i)\\
\hline
0 & A & 1/3 & 7\\
1 & XA & 2/3 \cdot 1/3 & 10 + 7 \\
2 & XXA & (2/3)^2 \cdot 1/3 & 20 + 7 \\
3 & XXXA & (2/3)^3 \cdot 1/3 & 30 + 7 \\
& ... & \\
i & (i \cdot X)A & (2/3)^i \cdot 1/3 & 10\cdot i + 7
\end{array}$$
and the expected time is:
$$\begin{array}{lcl}
E(T) & = & \sum_{i=0}^\infty P(i)T(i) \\
     & = & 1/3 \cdot \sum_{i=0}^\infty \left( 10 \cdot i  + 7 \right) \cdot(2/3)^i \\
     & = & 10/3 \cdot \sum_{i=0}^\infty i \cdot(2/3)^i 
       +    7/3 \cdot \sum_{i=0}^\infty (2/3)^i\\
     & = & 10/3 \cdot 6 + 7/3 \cdot 3 \\
     & = & 27 \\
\end{array}$$
A: Typical way to solve this:
The ant will either take path a and finish or take path b or c and be back in it's starting position.
Let $k$ be the number of times that the ant already has taken path b or c. Let $T_k$ be the expectation value for the time to finish for an ant that took already $k$ times the path. Then the expectation value for an ant with $k$ steps can be expressed in terms of an ant with $k+1$ steps.
$$T_k = \underbrace{\frac{1}{3} 7}_{\substack{\frac{1}{3}th\text{ chance to finish with path} \\ \text{ a in 7 minutes }}} + \underbrace{\frac{1}{3} (T_{k+1} + 8)}_{\substack{\frac{1}{3}th\text{ chance to finish with path b} \\ \text{ in 8 minutes} \\ \text{plus what ant $k+1$ needs on average }}} + \underbrace{\frac{1}{3} (T_{k+1} +12)}_{\substack{\frac{1}{3}th\text{ chance to finish with path c} \\ \text{ in 12 minutes}\\ \text{ plus what ant $k+1$ needs on average }}}$$
Since the ants are in the same starting position independent from the history (number of steps $k$) the average time is (you have $T_k = T_{k+1}$ which you can use to solve the above equation) :
$$T_k = \frac{1}{3}7+\frac{1}{3}(8+T_k)+\frac{1}{3}(12+T_k)$$
and after some rearrangments
$$T_k = 7+8+12 = 27 $$

Using an average
You can solve this with a mean, sort of.

*

*The ant finishes at least with path a which at least takes at least 7 minutes

*In addition, the ant has 2/3 probability to take paths b or c (each time) which take on average $\frac{8+12}{1+1} = 10$ minutes.

The mean times that the ant takes paths b or c is:
$$1 \cdot \frac{1}{3}\left(\frac{2}{3}\right) + 2 \cdot \frac{1}{3}\left(\frac{2}{3}\right)^2 + 3 \cdot \frac{1}{3}\left(\frac{2}{3}\right)^3 + 4 \cdot \frac{1}{3}\left(\frac{2}{3}\right)^4 + .... = \sum_{k=1}^\infty k \cdot \frac{1}{3}\left(\frac{2}{3}\right)^k = 2$$
note that this is related to a geometric distribution and the average number of extra steps that the ant needs is 2 (and each step takes on average 10 minutes). So the ant will take (on average):
$$ \text{ $7$ minutes $+$ 2 times $\times$ $10$ minutes  $= 27$ minutes}$$

Interestingly: you could also say the mean time for a single step is $9$ minutes (what you computed), and the mean number of steps is $3$, so the ant takes $3 \times 9 = 27$ minutes (you were not very far from the solution).
A: $$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$
$$T/3=9$$
$$T=27$$
From start point, each of 3 paths are equally possible. Two paths lead you back to start point.
A: Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment):
Both terms are geometric series:
$\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty i\cdot q^i=q \frac{d}{dq}\sum_{i=0}^\infty q^i=\frac{q}{(1-q)^2}, \text{ for } |q|<1$, so for $q=\frac{2}{3}$ this equals $6$.
$\sum_{i=0}^\infty (\frac{2}{3})^i$ is just a basic infinite geometric series and we have $\sum_{i=0}^\infty c\cdot q^i=\frac{c}{1-q}, \text{ with constant }c \text{ and } |q|<1$, which is $3$ for $q=\frac{2}{3}$ and $c=1$.
A: To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is,
$$
E[T] = \sum_{\text{path} \in \text{possible paths}} p(\text{path}) T(\text{path})
$$
Every possible path takes Passage $A$ only once, but can take passages $B$ and $C$ any number of times, in any permutation. So possible paths can be $A$, $BA$, $BBCCA$, $BCCBA$, etc. 
Suppose the probabilities of choosing passages $A,$ $B,$ and $C,$ are $p_a$, $p_b$, and $p_c$ respectively. Then, for example, the probability of taking path $CBBCCA$ is $p_a p_b^2 p_c^3.$ And, because there are ${5 \choose 2} = 10$ ways of taking $B$ twice and $C$ three times, the contribution of the expected path time given by the possibility of 3 $C$s and 2 $B$s is $10 p_a p_b^2 p_c^3 (T_a + 2 T_b + 3 T_c)$, where $T_a$, $T_b$, and $T_c$ are the path times of each passage respectively.
The above gives the contribution to the expected time for taking two $B$s and three $C$s before $A$. But in general, you have to sum over the expected time contribution of all possible combinations of paths $B$ and $C$ before $A$ I'll do that below, but I suggest you stop here and try it yourself first.

SPOILER
In general, the ant can take a non-$A$ passage any number of times between zero and infinity before taking passage $A$, and for that number of times, it can be any combination of passages $B$ and $C$. So, to get the expected path time, we sum up the contribution of all passage possibilities, multiplied by their time, which looks like,
$$
E[T] = T_a + p_a \sum_{n=0}^\infty \sum_{i=0}^n {n \choose i} p_b^i p_c^{n-i} [i T_b + (n-i) T_c].
$$
These sums can be evaluated. Using the Binomial theorem and taking the derivative, you can show that,
$$
\sum_{i=0}^n i {n \choose i} x^i y^{n-i} = n x (x+y)^{n-1}
$$
and
$$
\sum_{i=0}^n (n-i) {n \choose i} x^i y^{n-i} = n y (x+y)^{n-1}.
$$
Using these identities, we get
$$
E[T] = T_a + p_a (p_b T_b + p_c T_c) \sum_{n=0}^\infty n (p_b + p_c)^{n-1}.
$$
Taking the derivative of the sum of the famous geometric series, you can show that, for $|x| < 1$,
$$
\sum_{n=0}^\infty n x^{n-1} = \frac{1}{(1 - x)^2}.
$$
Noting that $1 - (p_b + p_c) = p_a$, we get,
$$
E[T] = T_a + \frac{1}{p_a} (T_b p_b + T_c p_c).
$$
If we take $p_a = p_b = p_c = 1/3$ and your values for the times, we get $E[T] = T_a + T_b + T_c$ which is 27 minutes.
A: It always take $7$ min to succeed, which happens once.
It takes an average of $10$ min to fail.
The probability of failing is $2/3$ on each try, or $(2/3)^n$ for n tries.
So the total time is ($7 + 10\cdot \sum_n (2/3)^n$) min.
Which is $7$ min + $2 \cdot 10$ min $ = 27$.
