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Former Member
Cruncher Joined: May 22, 2018 Post Count: 0 Status: Offline |
then what is the % chance that the event will happen at least one time?
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Former Member
Cruncher Joined: May 22, 2018 Post Count: 0 Status: Offline |
(1.01**100-1)%=1.74%?
----------------------------------------[Edit 2 times, last edit by Former Member at Sep 17, 2009 11:44:10 AM] |
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Former Member
Cruncher Joined: May 22, 2018 Post Count: 0 Status: Offline |
You do like this sort of thinks don“t you ..
----------------------------------------If there is a 1% chance something will happen, what are the chances it happens if you do it 100 times? The chance of not happening is P=(0.99)^100=36.60% The chance of happening at least once=1-P=63.40% You might be interested to know that there is a more general pattern here. If something has 1 chance in N of happening, what's the chance of it not happening after N times? For large N, the answer is e^(-1)=36.79%. Note that this value is very close to the actual value (36.60%) for N=100. If something has 1 chance in a 1000 of happening, the chance that it doesn't happen after 1000 times is 36.77% -- almost exactly the same as e^(-1). ("e" is the base of natural logarithms -- namely, 2.71828.... Also, e^(-1) is 1/e, or 0.36787...) There's more: What's the chance of it not happening after k*N times? Again for large N, the answer is e^(-k). (This is true for any non-negative value of k, fractional as well as integral.) Thus, if something has 1 chance in N of happening (for large N), the probability of it not happening after T tries is as follows: T=N prob=36.79% T=2N prob=13.53% T=3N prob=4.98% Example: Suppose you spin a big wheel marked with 1000 numbers, 1 to 1000. What's the chance that number 500 comes up? On the average, it appears once per 1000 spins. If you make 3000 spins, you expect this number to show up 3 times on the average, but there's a 5% chance that it doesn't show up at all. Example: Suppose a no-hitter occurs in one baseball game out of 1500, and that you attend all the home games (81) of your favorite team one season. What's the chance of seeing a no-hitter? In this example, N is 1500 and k=81/1500=0.054. The chance of a no-hitter is as follows: probability of *no* no-hitter = e^(-0.054)=94.74% probability of seeing at least one no-hitter = 1-94.74% = 5.26% Finally, instead of asking what is the chance of seeing one or more no-hitters, you might be curious about the chance of seeing exactly one, or exactly two. I won't go into the details here; but if you look up "Poisson distribution" on the web, you'll see how to calculate this. Here are the results (rounded to 0.01%): probability of exactly 0: 94.74% probability of exactly 1: 5.12% probability of exactly 2: 0.14% probability of exactly 3 or more: very small (I want to be clear that the Poisson distribution is strictly true only in the limit of large N. But for an example like a baseball no-hitter, where N=1500, the Poisson distribution is extremely close. If you want the exact answer, you need to use the binomial distribution, which you can also look up on the web. As N increases, the binomial distribution looks more and more like a Poisson distribution.) Finally, I'll get back to above (1% chance, 100 tries), and calculate the probability that the event will happen exactly M times. The exact probability according to the binomial distribution is prob = C(100,M) * (0.01^M) * (0.99^(100-M)) where C(N,M) is the combination function: C(N,M) = N! / (M! (N-M)!) Here are the probabilities (exact, but rounded to 0.01%): M=0: prob=36.60% (This is the result mentioned in the first paragraph of the answer.) M=1: prob=36.97% M=2: prob=18.49% M=3: prob=6.10% M=4: prob=1.49% M=5: prob=0.29% M=6: prob=0.05% M=7: prob=0.01% M=8 or more: prob < 0.001% The sum of the probabilities is 100%. The mean of this distribution is 1, and the mode is 1 (just barely, since the chance of 0 events is almost as much as the chance of 1). If I had used the Poisson distribution instead, the answers would have been close to these values, but slightly in error. The assumption throughout is that each trial is independent (e.g., you spin a wheel 100 times, and the outcome of one spin doesn't depend on the outcome of the previous ones). Once you get eaten by the shark, there are no more trials! Or maybe you have a large number of volunteers to replace the one who was eaten, but at that point the shark is no longer hungry. Period and out , no more posts here byeee ![]() [Edit 1 times, last edit by Former Member at Sep 17, 2009 12:41:44 PM] |
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Former Member
Cruncher Joined: May 22, 2018 Post Count: 0 Status: Offline |
![]() [Edit 1 times, last edit by Former Member at Sep 17, 2009 2:13:14 PM] |
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PecosRiverM
Veteran Cruncher The Great State of Texas Joined: Apr 27, 2007 Post Count: 1054 Status: Offline Project Badges:
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Okay now I've got a headache. I knew there was a reason I got a job where I didn't have to use that much of my brain.
---------------------------------------- Anyone seen my Advil?? ![]() |
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Sekerob
Ace Cruncher Joined: Jul 24, 2005 Post Count: 20043 Status: Offline |
An Anvil only gives more headache.
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WCG
Please help to make the Forums an enjoyable experience for All! |
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Former Member
Cruncher Joined: May 22, 2018 Post Count: 0 Status: Offline |
Wot's an advil?
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Sekerob
Ace Cruncher Joined: Jul 24, 2005 Post Count: 20043 Status: Offline |
It's an american headache pill containing ibuprofen. Now that one has zero pain killer effect on me, I mean zero.
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WCG
Please help to make the Forums an enjoyable experience for All! |
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PecosRiverM
Veteran Cruncher The Great State of Texas Joined: Apr 27, 2007 Post Count: 1054 Status: Offline Project Badges:
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Wot's an advil? Sorry cummerbund. It's a brand name like Sekerob said containing ibuprofen. I've had to switch from acetaminophen (Tylenol) due to dangerous side effects. ![]() |
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Former Member
Cruncher Joined: May 22, 2018 Post Count: 0 Status: Offline |
Excedrin. The best thing out there for my headaches. Have to watch out for the caffiene in it though.
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