# 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.

• You should write $\Omega=\{(a,b,c)\mid a,b,c\in\{S,F\}\}$. – drhab Mar 20 '18 at 14:22
• Yes sorry this was a mistake – user370967 Mar 20 '18 at 14:24

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:

$$P(A\cap B\cap C)=P(A)\times P(B)\times P(C)$$

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

$$P(A_1\cap A_2\cap A_3) = P(A_1)\cdot P(A_2)\cdot P(A_3) = \Big({1\over 3}\Big)^3$$

• I know it is correct. What are $A,B$ here explicitely? – user370967 Mar 20 '18 at 14:16
• I assume $A_1 = \{(F, b, c) : b, c \in \{S, F\}\}$. Similarly for $A_2$ and $A_3$ – Quoka Mar 20 '18 at 14:27
• True................ – Aqua Mar 20 '18 at 14:29