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A nuclear reactor is equipped with 2 safety systems. The first safety system will work in 90% of the cases. If it doesn't work, the second safety system will start to work, but it only works in 80% of the cases. What is the probability that no safety system will alarm the employers of the nuclear reactor?

My attempt:

Let $S$ denote that an alarm works and $F$ that no alarm works. Write $(F,S)$ for the outcome "The first alarm doesn't work, the second does" and analoguous for the others.

Then, $\mathbb{P}(\{F,F\}) = \mathbb{P}(\{(F,F),(S,F)\} \cap\{(F,F),(F,S)\}) $

$= \mathbb{P}(\{(F,F),(S,F)\} \mid\{(F,F),(F,S)\})\mathbb{P}(\{(F,F),(F,S)\}) = (0.2)(0.1) = 1/50$

Is this correct?

It seems tedious to write this always out this way. Can I avoid this?

  • "Assuming the events are independent, the probability that both alarms fail is $.2\cdot .1 = 1/50.$" – saulspatz Mar 20 '18 at 17:56
  • P(system 1 off and system 2 off) = P(2 off | 1 off) P(1 off), this means .2*.1 = .02. They give you the conditional probability: " the second safety system will start to work, but it only works in 80% of the cases." – yoshi Mar 20 '18 at 18:00

1 Answers1

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Yes, you can avoid it.

Let $F_1$ denote the event that the first alarmsystem will fail and let $F_2$ denote the event that the second alarmsystem will fail.

Then: $$\mathsf P(F_1\cap F_2)=P(F_1)P(F_2\mid F_1)=P(F_1)P(F_2)=0.1\cdot0.2=0.02$$here the second equality is based on independence.

Note that this can apparantly be solved without any bothering what the underlying probability space looks like.

Here you find a nice question about that subject.

It is important that you know about underlying probability spaces but there comes a time (if you persevere in probability theory) that you will hardly ever bother about them anymore.

drhab
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