Probability of Independent EventsDate: 02/07/97 at 18:36:51 From: Carl J. LaGrassa Subject: Probability question Given that the probability of an event is known, is there a formula which can give the probability of an event in a given number of trials? To use specific examples, I am interested in the probability of an event occurring in 16k, 32k, and 64k when the probability of occurrence is 1/32000. Thanks for your consideration. Date: 02/08/97 at 05:31:03 From: Doctor Mitteldorf Subject: Re: Probability question Dear Carl, There aren't many formulas for probability questions. The big question here is whether your 16k copies of this problem are all independent of one another. Does the occurrence of the event in one place forebode an enhanced probability that it will appear in another place? With that caveat, I can give you a formula: for independent events, each with probability p, the probability of n such events is: 1 - (1-p)^n For your first example, this is 1-(31999/32000)^16000. Think of it this way: The first one has a (31999/32000) chance of not failing. If it doesn't fail, the second one has a (31999/32000) chance, etc. If p is small, this turns out to be approximately equal to e to the power (-n/p). - Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 02/08/97 at 22:02:25 From: Carl J. LaGrassa Subject: Re: Probability question I'm afraid I don't understand the answer. I was speaking of independent events. I do not understand the notation "^" or the reference to "e". Specifically, in video poker a royal flush will occur randomly with a frequency of 1/32000 hands played. I am looking for the probability of getting an occurrence in a given sample size. Date: 02/09/97 at 06:21:46 From: Doctor Mitteldorf Subject: Re: Probability question Dear Carl, The symbol ^ is used on computers to indicate "to the power of". x^2 means x squared. I use it here because it's not easy to send raised exponents over the internet. The point is simply that if two events are independent, you can multiply their separate probabilities to get the probability of both happening. If the probability of a single hand coming out not-a- royal-flush is 31999/32000, then the probability of two coming out not-a-royal-flush is (31999/32000)^2 and the probability of three coming out not-a-royal-flush is (31999/32000)^3. To get the probability that none of the 16000 trials will come out a royal flush, simply multiply sixteen thousand times the probability that each separately will come out not-a-royal-flush. This can be written (31999/32000)^16000. Take the number (31999/32000) and multiply it by itself 16000 times. The statement that there will be at least one royal flush in the 16000 trials is just the opposite of the statement "none of the trials will produce a royal flush." So you can get that probability by subtracting that product from 1. -------------------------------------------- e is the base of natural logarithms and is approximately equal to 2.71828182845... One definition of e is that if you take larger and larger numbers n and calculate the nth power of (1+1/n), the answers come out closer and closer to e. This is something you can try on a calculator. It also works that the nth power of (1-1/n) comes closer and closer to 1/e as n gets larger and larger. Try it! You can use this last fact to come up with an approximate expression for (31999/32000)^(16000). The thing you're raising to a power, (31999/32000), is very close to 1. Think of it as 1-1/n, where n = 32000. You're not raising it to the nth power, but to half n. We use the fact that anything to the power of half n is just the square root of the thing raised to the nth power. (Try proving that by multiplying two copies of the thing together, and simplifying what you get.) So (31999/32000)^16000 is very close to e^-0.5, or 1/e raised to the power 1/2. Thanks for writing back. Please let me know if this is clear now. -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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