Assume that Bob carries eggs from point $A$ to $B$. He can carry $1$ egg each time. Let the probability that Bob breaks an egg be $0.99999$ which is almost a certain event (for me). If Bob carries $100$ eggs separately, can we say the probability of Bob breaking an egg is $0.99999 \% = 0.00999$?

I am asking because $0.99999$ is a very high probability in my opinion, and changing the try count doesn't have any practical effect on the above example.

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    $\begingroup$ I have edited the title. This is a good question, but the title will be misleading for many people, as seen by some of the answers below. That is because there is a concept in probability called an "almost certain" event, unrelated to the actual question. Ricardo, in the future, it might be safer to use the phrase "very probable" instead. $\endgroup$
    – Garrett
    Feb 14 '14 at 10:31
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    $\begingroup$ What exactly is the improbable event in the title? The body only mentions highly probable events. (What seems improbable is that anybody would allow Bob carrying their eggs.) $\endgroup$ Feb 14 '14 at 11:40
  • $\begingroup$ @MarcvanLeeuwen, you're right, my edit to the title was confusing. I've edited it again. $\endgroup$
    – Garrett
    Feb 16 '14 at 1:20

The probability that Bob breaking an egg is $1$ minus the probability that he breaks no eggs. For each of 100 trips, the probability of him not breaking the egg is $0.00001$. Can you conclude?

  • $\begingroup$ Sounds credible. So chance is zero point 401 nines... $\endgroup$
    – nl-x
    Feb 13 '14 at 21:00

this is the probability that bob breaks at least one egg if he carries 100 eggs separately $1-(1-0.99999)^{100}$

  • $\begingroup$ So if he took just 1 egg, propability of breaking at least an egg is 1-0.99999^1 ? Not $\endgroup$
    – nl-x
    Feb 13 '14 at 20:58
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    $\begingroup$ @nl-x, no, it would be $1 - (1 - 0.99999)^1$ which is $0.99999$ as defined. You can read the formula as "the probability that not none of the eggs are broken." $\endgroup$
    – wchargin
    Feb 14 '14 at 0:12
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    $\begingroup$ which is 0.999... (with five hundred 9's total) $\endgroup$ Feb 14 '14 at 1:37
  • $\begingroup$ @WChargin Yes, it would be..AFTER the anwer was EDITED ! math.stackexchange.com/posts/675513/revisions $\endgroup$
    – nl-x
    Feb 14 '14 at 6:43
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    $\begingroup$ I cant believe someone breaks egg at 0.99999 probability each time. And he is still trusted with handling 100 eggs. $\endgroup$
    – lulu
    Feb 14 '14 at 18:37

In probability theory, impossible events have probability $0$, but an event having probability $0$ does not guaranty it will not occur.
So a "proper" almost impossible event, according to the theory, will have probability $0$.

Bob, as terrible as he his in holding eggs while translating in space, might surprise everyone and keep delivering intact eggs one after the other.. after the other.. after the other....
(If you dare arming him with a non depleting arsenal of eggs)

  • $\begingroup$ @Dror Is this an example of your theory: What is the probability that three random points on a circle will connect to form a right triangle? $\endgroup$
    – Sawarnik
    Feb 14 '14 at 12:40
  • $\begingroup$ @Sawarnik It's not my theory. Pickup a random number from any real interval, and derive the probability of it coming out irrational (It'll be $1$). Then look at the probability of the complement event, $0$. So does this mean rational numbers don't exist...? (No) $\endgroup$
    – user76568
    Feb 14 '14 at 14:34
  • $\begingroup$ Ok. But does my example also fit the bill? $\endgroup$
    – Sawarnik
    Feb 14 '14 at 14:37
  • $\begingroup$ @Sawarnik I think so, yes. $\endgroup$
    – user76568
    Feb 14 '14 at 14:45

Just to make one more observation about "almost impossible events". Consider the experiment of picking a random real number in the interval I := [0, 1) (say, but it's not really important, using an uniform random distribution) and the question "How probable is it that the picked number is rational"?

The set of non-rational numbers in I is uncountable, whereas the set of rational numbers in the same interval is countable. It means that the probability of picking a non-rational number is infinitely larger than that of picking a rational number and, consequently, the probability of picking a rational number is 0. However, nothing prevents you from picking 0.5, which is a perfectly possible result of your experiment.

As Dror points out, some events can have probability 0 and this does not mean that there is no possibility for them to occur.

What is the probability of an almost impossible event?

Mathematically speaking, it is 0.

"Almost" has a well-defined mathematical meaning. A function can be almost everywhere continuous, meaning that it is continuous everywhere but in a countable set of places. The Dirac Delta function is almost everywhere zero.

EDIT: no, the explanation of "almost everywhere continuous" is not the one given in the last paragraph. The truth is more complex than this, and the linked Wikipedia article gives a correct explanation.

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    $\begingroup$ Being almost everywhere continuous does not mean one is "continuous everywhere but in a countable set of places". $\endgroup$
    – Did
    Feb 14 '14 at 8:39
  • $\begingroup$ @Did I made that mistake my self once(?).. I think It's a common story being told to students in courses preceding measure theory, and we hold on to that story.. ;-) $\endgroup$
    – user76568
    Feb 14 '14 at 15:01
  • $\begingroup$ Uhmmm... You are right, guys. I should probably remove that final part of my answer, which is not even directly related to the original question. I let it in, since otherwise this discussion in the comments makes no sense. @Dror: or it's probably a story told to non-math students by non-math professors? It is my case (I'm a computational scientist). $\endgroup$
    – Spiros
    Feb 14 '14 at 15:09
  • $\begingroup$ I study math, and I can say from my experience that I was taught things which are not correct. In most of the cases, teachers in following courses make it very clear that It's not correct. In other cases I recognize the mistakes I'm told my self and go "Whaaat..?" (In my head). That's reality for you. (Or just for me) $\endgroup$
    – user76568
    Feb 14 '14 at 15:18
  • $\begingroup$ Yes, such things happen almost everywhere. In my case, I can remember very well the day our professor for computational physics told us that the Jacobi iteration converges to the solution for all matrices, unconditionally. This is the case we where like "Whaaat!" (not just in our heads). Anyway this is going off-topic. I'll update the answer to make it clear that there is something wrong in it. $\endgroup$
    – Spiros
    Feb 14 '14 at 23:46

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