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Suppose there are 2 points A and B. We can go from A to B through 2 paths, path 1 and path 2.

Path 1 has 0.9 probability of being blocked out (unable to use path 1 to go from A to B). Path 2 has 0.8 probability of of being blocked out.

What is the probability of success in going from A to B?

Since we have 2 alternatives, that is path 1 or path 2, P(A to B) = 0.9+0.8 = 1.7. What is the flaw in getting probability > 1?

I believe the answer is 1 - 0.1*0.2, but I don't know what is wrong with 0.9+0.8.

Can someone please clarify?

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    $\begingroup$ Separate from your question, but are you mixing blocked and open? P(path 1 blocked)=0.9, but then you appear to have P(A to B on path 1)=0.9. That doesn't help explain what's wrong with 1.7, but swapping "unable" with "used" doesn't help. $\endgroup$ Commented Dec 16, 2024 at 19:57
  • $\begingroup$ You could start with a "casuistic" approach (simply list all possible combinations, which is for such small problems entirely doable) and arrange them in a visual grid. with e.g. color codes. That would show, Venn-diagram like, how the single probabilities interact and combine into an overall probability. $\endgroup$ Commented Dec 17, 2024 at 15:33

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The rule of addition - $P(X\text{ or } Y)=P(X)+P(Y)$ only applies if the events are disjoint, that is, they can't both happen. Here clearly the paths can both be available.

In general, the rule becomes $P(X\text{ or } Y)=P(X)+P(Y)-P(X\text{ and } Y)$. This is because adding the probabilities covers the times when both paths are open twice, so you need to subtract that off to only count it once.

If the events you wanted had probability $0.9$ and $0.8$, this would give $0.9+0.8-0.9\times 0.8=0.98$, as expected. However, note that in this question $0.9$ and $0.8$ are not the probabilities of the paths being available, but the probability of them being unavailable.

Therefore you actually want $0.1+0.2-0.1\times 0.2=0.28$.

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    $\begingroup$ Another way to get to 0.28 would be 1-0.9*0.8. $\endgroup$ Commented Dec 16, 2024 at 20:04
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    $\begingroup$ Since the question asked ‘What is the flaw in getting probability > 1?’, it might be worth emphasising that all of the (correct) probabilities are indeed less than $1$. $\endgroup$ Commented Dec 17, 2024 at 9:58
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This might be a more intuitive way to look at it.
All in all, there are 4 distinct possibilities: A and B are both available, only A is available, only B is available, neither of them are available. All of these possibilities have their own probability, and since these are the only possible outcomes and no two outcomes can happen at the same time, their probabilities must add up to 1.
The question here asks for the sum of the first three possibilities, we see now that we can calculate this as 1 minus the probability that A and B are both NOT open. That's how you get to the 1-0.9*0.8=0.28.
So this means that, if you could check both paths before you start, you have a 28% chance that at least one of the paths is open.
If, however, you randomly choose a path first (both paths equally probable), and only then check whether it's available or not, then the probability of the path being blocked is $0.5*.9+0.5*0.8=0.85$. You will have a 85% chance that your randomly chosen path will be blocked, or 15% chance that it's open.

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  • $\begingroup$ Your edit was good; I removed my downvote. But I haven’t upvoted yet, because I think you still need to work on clarity. You have many different descriptions here (‘available’, ‘open’, ‘blocked’ and, perhaps most confusingly, ‘NOT open’). Also, you say things like ‘if you could check …, you have a 28% chance that at least one of the paths is open’. No, you have a 28% chance of at least one path being open, period. What you mean is that you have a 28% chance of success, taking into account both the way you select a path and whether the paths are open. $\endgroup$ Commented Dec 19, 2024 at 6:23
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Well, it depends. Given

  • probability for path 1 to be successful $= 10$%
  • probability for path 2 to be successful $= 20$%

Then, the probability for any path to be successful (assume, you first try one and if not successful, you try the other) is $>= 20$% and $<=30$%; the exact value depends on a potential correlation of the 2 paths to be blocked.

If the 2 paths being blocked is independent, then the probability for success is

  • First trying one, then the other $10$% $+90$% $20$%$= 28$%
  • First trying the other, then the one $20$%$ + 80$% $10$%$ = 28$%
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