I need someone who knows lots of math?

I need someone who knows lots of math? Topic: Define use case scenario defined
June 16, 2019 / By Hezron
Question: do you see any mistakes in this? lets start with the probability of an event (any event) occuring, we will call this probability p then, it follows, that the probability of this event NOT occuring is (1-p) probabilities of independent events multiply, so the chance that this event does not occur in N independent trials is (1-p)^n and, then the probability that this event WILL occur in at least one of those N trials is 1-(1-p)^n the interesting part comes now, when we set the probability of the event occuring to ZERO, but give it an infinite number of chances. logic would tell you that if an event has zero probability-absoulutely no chance of ever occuring, then it will never happen, no matter how long you wait. it is impossible. or is it? Using the probability equations found above to calculate this strange scenario, you need to take the limit of the expression as p approaches zero and n approaches infinity. the equation becomes n lim lim 1-(1-p) p->0 n->inf. infinity can be defined as the limit of 1/p as p approaches zero, so this is equivalency to 1/p lim 1-(1-p) p->0 if a new variable, m, is introduced, and defined as m= - p then the equation is transformed to -1/m lim 1-(1 m) - m->0 since zero is neither positive or negative, this can be rewritten -1/m lim 1-(1 m) m->0 introducing yet another variable, q, and defiining it as q=1/m the equation takes on the form -q lim 1-(1 1/q) 1/q->0 recognizing that as 1/q -> 0, then q must tend to infinity, and that minus q is equivalent to q times a negative one, this can be rewritten as (-1)q lim 1-(1 1/q) q->inf the laws of limits allow me to move the 1 to the outside of the limit, and the exponent laws let me break up a^(-q) into (a^q)^(-1). this simplifies the problem as follows: q -1 1- (lim (1 1/q) ) q->inf the quantity q lim (1 1/q) q->inf is quite well known in calculus as the limit that equals the number e. taking this into consideration, this problem is simplified immensely. q -1 1- (lim (1 1/q) ) is transformed into q->inf -1 1- e or, simply, 1-1/e if an approximate value is taken, 1-1/e equals 0.632, or 63.2% ok, so now, what does all of this mathematical-crap actually mean??? think about it this way. the original question was "What is the chance that event A will occur, given that its probability of happening is zero, but it is given infinitely many tries?" This is quite like asking "if I have a quarter, and I flip it an infinity of times, what is the chance that one of those times it will turn into a fork?" the logical answer to this question is "it will never turn into a fork. Its a quarter. duh" but as I showed in the mathematics above, there is indeed a chance that one time out of those infinity of tries, the quarter will turn into a fork. and this chance is greater than one half. 63.2 percent, to be precise. i realized that i did not explain the probabilities very good with words. the 63% chance is the chance that the quarter will turn into a fork ONE time out of the infinity of tries. that is, if you took 100 quarters, and flipped them an infinity of times, 63 of them would eventually turn into a fork, or something equally bizarre. i just noticed, that the "plus" signs are not showing up..... there should be a plus sign in between the 1 and the m&1/q's throughout the proof
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Best Answers: I need someone who knows lots of math?

Elmer Elmer | 9 days ago
Egad... your proof is bad, and your abuse of notation is worse. Is it really so hard to proofread the stuff you copy and paste BEFORE posting it? Mistake #1: "logic would tell you that if an event has zero probability-absoulutely no chance of ever occuring, then it will never happen, no matter how long you wait. it is impossible." FALSE. Although all impossible events have a probability of zero it is not the case that all events with a probability of zero are impossible. Let S be any finite subset of the natural numbers. Select a natural number at random. The probability that it will be a member of set S=(# of objects in set s)/(# of natural numbers)=N/∞=0, but the set S is nonempty, so it is clearly possible that the number selected is a member of S, even though the probability is zero. Mistake #2: "infinity can be defined as the limit of 1/p as p approaches zero" FALSE. There is no limit of 1/p as p→0. In order for a limit to exist, p has to approach the same limit from both the right and left hand sides, but the LHL of 1/p as p→0 is -∞. Thus, the definition of infinity is wrong. Mistake #3: "since zero is neither positive or negative, this can be rewritten... m->0" This doesn't work, because of mistake #2. While you could potentially salvage mistake #2 by defining it specifically as the RHL of p, doing so would later prevent this particular abuse of signs. The remainder of the proof is actually fine, IF my guess as to where your misplaced exponents are supposed to be is correct. Actually, you can get the same result very simply as follows: P=(n→∞, p→0)lim 1-(1-p)^n P=(n→∞, p→∞)lim 1-(1+(-1)/p)^n P=(n→∞)lim 1-(1+(-1)/n)^n Note that (n→∞)lim (1+x/n)^n=e^x, so since x=-1 P=1-e^(-1) P=1-1/e≈0.63212 So it would seem that you managed to get a "correct" result even though your methods were invalid. Of course, that would imply that any probability zero event has the same odds of occurring at least once in an infinite number of trials, regardless of what that event is, or whether it is impossible or merely infinitely improbable. So there is a flaw, and now that I have simplified the proof, it is easy to see where it lies: Mistake #0: "P=(n→∞, p→∞)lim 1-(1+(-1)/p)^n therefore P=(n→∞)lim 1-(1+(-1)/n)^n" This is false. Consider the following: (p→0, q→0)lim p/q. Making that substitution would imply that the limit as we approach 0/0 is always 1. However, if we let p=10x and q=x, then this becomes (x→0)lim 10x/x, which is obviously 10. This form is indeterminate - we cannot evaluate this limit simply by substituting the limits of the individual parts, nor by assuming the parts are equal. 1^∞ is another indeterminate form: lim (p→1, q→∞)lim p^q is free to take on any form depending on how the functions p and q behave, and it is not in general permissable to assume that they are related. If we assume that an event is impossible for instance, then the probability of the event is zero, regardless of the value of n, and so the correct limit would be (n→∞)lim 1-(1-0)^n, and this is clearly zero, which is what we would expect, given that the event is impossible. Now, when does this limit apply? It applies if and only if the probability of an event on any given trial is inversely proportional to the number of trials. For instance, suppose we play a game where 1 blue ball and n-1 red balls are shuffled into a bag, and we get to pick from the bag n times (with replacement). We win the game by picking the blue ball. In this case, our odds of winning are given by P=1-(1-1/n)^n, and our odds of winning this game if we play with a countably infinite number of balls really are 1-1/e, even though the probability of winning on any particular draw is zero. But these kinds of situations are exceptions rather than the rule, and in general the probability of an event happening on any given trial is unrelated to the number of trials, in which case the probability of a probability zero event happening at least once in an infinite number of trials is still zero. Edit: Although, ironically, this doesn't disprove the idea that a quarter will turn into a fork if flipped an infinite number of times. Quantum mechanics teaches us that any atomic particle does not have a definite position and velocity, but rather has some probability of having any given position and velocity, and thus has some finite (though vanishingly small) probability of being found anywhere in the universe (or at least within the light cone of the last observed position). Therefore, there is is some finite probability of an atom in a quarter being found at its corresponding position in a fork, and since there are a finite number of atoms in a quarter, it follows that the probability of a quarter spontaneously turning into a fork is finite and nonzero (although so small that we would require the inverse of a power tower to describe it efficiently). Thus, from the infinite monkey theorem, it follows that the probability of a quarter turning into a fork at least once while being flipped an infinite number of times is 1 - that is, it will almost surely happen. Just try wrapping your mind around that for a while.
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Elmer Originally Answered: HELPP i need lots and lots of help pleaase?
you might want to put this question in a different section to start. and just put together what you just said into an actual sentence!

Clancey Clancey
I think your question is about interpreting the answer for what happens when an event that has zero probability of occurring is attempted infinitely many times... The short answer is that it cannot occur. If the probability of an even occurring is zero, then it's not going to happen. You should be able to determine how this differs from what happens when the probability of an answer occurring approaches zero. As you have seen from taking your limits and obtaining an answer realted to e, the cumulative effect of an infinite number of tries suggests that it has approximately a 63% of EVER happening. Think about it like this: you walk out the back door of your house and set foot on a concrete step. The step isn't going to break from you stepping on it once. But what if you stepped on it an infinite number of times? In that situation, the event has some possibility of occurring, but allow me to use your example. You toss a coin into the air and expect that it will generate a result. Based on this logic, you could say that P(getting a result from a coin toss) = 1, but that will depend on what your definition of a result is. One could suggest that tossing a coin has the POTENTIAL of not giving either heads or tails because it landed on it's side. This is VERY unlikely, but given an infinite number of attempts, it is still possible. I just hope I have understood your question...
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Amittai Amittai
I think that you are wrong in trying to set specific odds of occurrence using the limits as a value approaches infinity. However, I think that you are correct that a zero-probability event may occur in an infinite amount of trials. Perhaps in some sense this could explain the big bang. Within our physical observable universe, an infinite number of trials is probably not feasible, because there is a minimal plank time and minimal plank distance, so there are only a finite number of events that can occur within a finite volume of space over a finite time. So this is more of probably more of theoretical than a practical discussion.
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Tilly Tilly
Holy long question, lookin at the original equation however I'd say that if the probability is 0, and you give it an infinite number of times to happen, it will happen once, because 1-0=1 and 1 to the power of anything is still 1, so it has a once in infinity chance of happening, which is the best way to say it because saying something will never happen is ludicrous because we don't even know all the laws in the universe yet, they may be different in different areas.
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Roxie Roxie
The two that stood out to me were: If you define p as 0, then it makes no since to take the limit as p->0 of 1-(1-p)^n. Set p=0-- 1-(1-0)^n=1-1^n=1-1=0, independent of n. Also, somewhere in there you say that the limit of (1-1/q)^q as q-> infinity =e. If you write 1-1/q out in a slightly less simplified form, you get (q-1)/q. it's easy to see as q goes to infinity, this goes to 1. So, you end up again with 1-1^(very large number), which is 1-1=0
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Natalee Natalee
You lost me here: n lim lim 1-(1-p) p->0 n->inf. infinity can be defined as the limit of 1/p as p approaches zero, so this is equivalency to 1/p lim 1-(1-p) p->0 ... How is n = 1/p ???
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Lorna Lorna
the probability of an event occuring in N random trials is not 1-(1-p)^n, but rather, it is p^n, so your initial assumption is faulty
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Lorna Originally Answered: Confused About What to Do?(Lots of info sorry if tmi)?
It's always important to follow your medical advice from your doctor. If your unsure I would consult him to see what the best choice is for you and your body. Every woman is different so it's really hard for a complete stranger to give medical advice without knowing your medical history. Now the "Pull out" method is not the safest birth control method out there. All it takes is one sperm to implant in that egg and next thing your know, your pregnant. If I were you I would continue using condoms with your partner to eliminate the possibility that you may get pregnant. Also I'm sure when you ask your doctor he will tell you that switching from one form of birth control to another may increase the chances of pregnancy anyway. So yeah, stick with the condom just in case hun. Good luck.

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