Another probability question 
Suppose we perform a series of consecutive experiments, where the
  outcome of one experiment does not affect the outcome of another
  experiment. Suppose that there is a probability of $1/3$ that an
  experiment fails.
We perform $3$ consecutive experiments. What is the probability that
  all three experiments fail?

We work in the sample space $\Omega:= \{S,F\}^3 = \{(a,b,c)|a,b,c \in \{S,F\}\}$
where $S$ denotes succes and $F$ denotes failure of the experiment.
Then, $\mathbb{P}(\{(F,F,F)\}) = \mathbb{P}(F)^3 = 1/27$
but I'm unsure why I can formally perform this step? We have to keep working in the same probability space. 
I do know that this question is a special case of Bernoulli experiment but let's ignore that for the sake of the question.
 A: Yes, your solution is correct. If two events are independent we have $$P(A\cap B) = P(A)\cdot P(B)$$
So if $A_1,A_2,$ and $A_3$ are consecutive results, we have 
$$P(A_1\cap A_2\cap A_3) = P(A_1)\cdot P(A_2)\cdot P(A_3) = \Big({1\over 3}\Big)^3$$
A: From this website, it seems like the rule of conduct applies:

If $A$ and $B$ are independent events, then:
$P(A\cap B)=P(A)\times P(B)$

For this specific problem, "the outcome of one experiment does not affect the outcome of another experiment", so the experiments are independent events, you can apply the rule of conduct above, so you are doing it right.
Note that the rule above can be applied for three events instead of tw, or even $n$ events that are independent.
$P(X)$ is the probability of the event $X$ happens (or event $X$ is true), in this case, there are three events. Let $A$ be the event "first experiment fail", $B$ be the event "second experiment fail", $C$ be the event "third experiment fail", then we will have:
$$P(A\cap B\cap C)=P(A)\times P(B)\times P(C)$$
