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I'm new to statistics and probability. I came upon this problem while solving Erwin Kreyszig's book. The problem is given like:

A motor drives an electric generator. During a 30 days period, the motor needs repair with 8% and the generator needs repair with probability 4%.What is the probability that during a given period, the entire apparatus(consisting of a motor and a generator) will need repair?

So, $\mathsf P(\text{motor-repair}) = 0.08$ and $\mathsf P(\text{generator-repair}) = 0.04$.

Now why $\mathsf P(\text{motor-repair})\cdot\mathsf P(\text{generator-repair}) = 0.04\cdot 0.08$ should not give the probability of the entire apparatus needing repair?

The solution is given like $$\mathsf P(\text{motor-repair}) + \mathsf P(\text{generator-repair}) - \mathsf P(\text{motor-repair}) \cdot\mathsf P(\text{generator-repair})$$

Also in another resource it is given like $$1 - ((1-\mathsf P(\text{motor-repair}))\cdot (1−\mathsf P(\text{generator-repair})))$$

My question is how these solutions are coming?

When should I multiply and when should I add in Probability?

What is the difference between $\mathsf P(A)\cdot\mathsf P(B)$ and $\mathsf P(A \cap B)$?

I'm sorry if I'm asking stupid questions, but these things are not becoming clear. Please help me. Thanks in advance.

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3 Answers 3

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Add for (disjoint) unions, multiply for (independent) intersections.

Roughly speaking (not always 100% true!), in probability, the word or translates into addition, while and translates into multiplication.

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  • $\begingroup$ Okay! I got it now.. That means like @Anirudh pointed out they want me to find the union and since it is independent like motor has nothing to do with generator P(A∩B) is becoming P(A).P(B)? $\endgroup$
    – lu5er
    Jul 1, 2017 at 5:37
  • $\begingroup$ Actually I would suggest you please don't look into the way how they have approached the problem... Try to build a solution first yourself... It really helped me because I was very weak at Probability around a year ago... $\endgroup$
    – Aditya
    Jul 1, 2017 at 5:42
  • $\begingroup$ Actually I did that. But it makes me mad when I see none of my methods work. $\endgroup$
    – lu5er
    Jul 1, 2017 at 5:43
  • $\begingroup$ But one thing - "consisting of a motor and a generator" then shouldn't it be like P(A∩B) like you said? $\endgroup$
    – lu5er
    Jul 1, 2017 at 5:45
  • $\begingroup$ Can you refer the page number.. $\endgroup$
    – Aditya
    Jul 1, 2017 at 5:45
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Now why $\mathsf P(\text{motor-repair})\cdot\mathsf P(\text{generator-repair}) = 0.04\cdot 0.08$ should not give the probability of the entire apparatus needing repair?

Why, yes, for a certain interpretation of "entire apparatus needs repair"; this just does not appear to be the intended meaning.

Assuming the components fail independently, that is the probability that both need repair.   The probability for the intersection of (pairwise) independent events equals the product of the probabilities for each event.

Note: The independence of the events is important here.   It is indeed definitional. $$A\perp B ~\iff~ \mathsf P(A\cap B)=\mathsf P(A)\cdot\mathsf P(B)$$

The solution is given like $$\mathsf P(\text{motor-repair}) + \mathsf P(\text{generator-repair}) - \mathsf P(\text{motor-repair}) \cdot\mathsf P(\text{generator-repair})$$

That would be the probability that either component needs repair.   It is apparent that they intended that the entire apparatus needs repair means that either part does.

Now, the probability for the union of disjoint events equals the sum of the probabilities of the events.   However, these events are not disjoint, so we apply the Principle of Inclusion and Exclusion.

$$\mathsf P(A\cup B)~{=\mathsf P(A\cup (B\setminus A))\\=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B) \\ = \mathsf P(A)+\mathsf P(B)-\mathsf P(A)\cdot\mathsf P(B)\qquad\text{if independent}}$$

Also in another resource it is given like $$1 - ((1-\mathsf P(\text{motor-repair}))\cdot (1−\mathsf P(\text{generator-repair})))$$

That would be the probability that both components do not require repair. (ie: That neither does.)   Which is the same thing thanks to deMorgan's Laws.$$(A\cup B) ~=~ {(A^\complement\cap B^\complement)}^\complement$$

So therefore, since the complements of the events are also independent from each other:$$\mathsf P(A\cup B)~{=~1-\mathsf P(A^\complement\cap B^\complement)\\ = ~\bbox[pink]{ 1-(1-\mathsf P(A))\cdot(1-\mathsf P(B)) }\\ = ~1-(1-\mathsf P(A)-\mathsf P(B)+\mathsf P(A)\cdot\mathsf P(B))\\=~\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)}$$

Which is indeed a proof of the Principle of Inclusion and Exclusion holding for two independent events.

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  • $\begingroup$ And this completes my question. Thank you very much. :) $\endgroup$
    – lu5er
    Jul 1, 2017 at 16:41
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From what you said of their intended solution, I think they meant that if either of the parts is broken the entire apparatus needs repair.

Your solution is calculating the probability that both parts in the apparatus need repair, so I think you may have misread the question a bit.

If you are confused on when you should add and subtract in probability I would try this site here

If you still don't understand the whole addition and multiplication thing leave me a comment so I can go more in depth on my own later.

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  • $\begingroup$ Thanks for the reply! But then P(AUB) = P(A) + P(B) - P(A∩B). So, P(A∩B) = P(A).P(B)? Can you help? $\endgroup$
    – lu5er
    Jul 1, 2017 at 5:34
  • $\begingroup$ Your values are correct but you are getting them mixed up you are looking for A and B but they are looking for A or B $\endgroup$
    – Anirudh
    Jul 1, 2017 at 5:38
  • $\begingroup$ In case ur not clear on some of the syntax $\endgroup$
    – Anirudh
    Jul 1, 2017 at 5:39
  • $\begingroup$ Can you say something on the second solution? $\endgroup$
    – lu5er
    Jul 1, 2017 at 5:40
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    $\begingroup$ P(A and B)=P(A) *P(B) in this case and this becomes important later when we want to calculate P(A or B) P(A or B)=P(A) +P(B) seems like a logical solution but since A and B are not mutually exclusive events we have just overcounted by P(A and B) therefore we get P(A or B) =P(A) +P(B)-P(A and B) $\endgroup$
    – Anirudh
    Jul 1, 2017 at 5:49

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