The random variable has a Poisson distribution with parameter L = 1*50/20 = 2.5.
So Prob(at least one event in 50 years) = 1 - Prob(No events)
= 1 - L0e-L/0! = 1 - e-2.5 = 0.918 approx.
The probability of an event is one when the event is absolutely certain to happen. It is more useful to understand why the sum of probabilities of an event adds up to one. If you have an event with a probability of p, then the probability of the event not occurring is 1 - p. This is because p plus (1 - p) is 1. That seems trivial, but consider a realistic example... Lets say you are subject to random drug testing, and that you are in a 100% pool such as might be mandated by 10CFR26, "Fitness for Duty in a Commericial Nuclear Power Plant". What is the probability that you will be chosen at least once in a year? Well, if sampling occurs five days a week, then there are 260 samples per year. The probability of being chosen in one sample is 1 in 260, or 0.003846. That means the probability of not being chosen is 259 in 260, or 0.9962. The probability, then, of not being chosen in one year is (259 in 260) raised to the 260th power, or 0.3672, which is about 95 in 260. Invert once again, and you see that the probability of being chosen at least one time in a year is 165 in 260, or 0.6328.
If at least three flips are tails, there are two scenarios where we can obtain exactly four tails in five flips. Either the first four flips are tails and the last flip is heads, or the first flip is heads and the next four flips are tails. Each scenario has a probability of 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32. Therefore, the probability of obtaining exactly four tails in five flips if at least three are tails is 1/32 + 1/32 = 1/16.
The probability of landing on heads at least once is 1 - (1/2)100 = 1 - 7.9*10-31 which is extremely close to 1: that is, the event is virtually a certainty.
Depends on the probability of reading any.
Not knowing exactly what the 'increase' is, it is hard to answer this. Here is what happens if the event probability is the same each time (at the bottom is one example where probability increases):Let p = probability that the event will occur on a trial (attempt).Let q = probability against the event occurring on a trial = 1 - p. [100% - probability for]Probability that the event will not occur on n trials = qn, so the probability that the event will occur at least once on n trials is 1 - qn = 1 - (1-p)n.Example: Say you roll a 6-sided die twice and want the probability that you'll get a 1 at least one of the throws.p = 1/6, n = 2 : Probability = 1 - (1 - 1/6)² = 30.56%Let's look at another way to figure this. Rolling a die twice, it's easy enough to make a 2-dimensional table of the results. There are 36 possibilities, and 11 of them have a 1 occurring. So the probability is 11/36 = 30.56%If you were to do 3 attempts, you would have a 3-dimensional table with 216 possibilities, and 91 of them have a 1 occurring at least once.91/216 = 42.13%, which is the same using the formula: 1 - (1 - 1/6)³ = 42.13%So for the problem where the probability increases each time. As an example, say you're playing BINGO with 75 balls. Say you're playing a blackout game and 50 balls have already been drawn, and you've filled your block except for a single square (say it's number 1). So there are 25 balls remaining. It will be similar to above, except you cannot just square or cube the probabilities because they change.You still want to calculate the probabilities against, then multiply them together then subtract from 100%. Since this is an example, you're playing BINGO and you look around at several neighbors and see that everybody around you has three spaces to fill, so you want to hit #1 in the next three balls.Assign the probabilities for picking #1 on each trial as p1, p2, p3.So p1 = 1/25. If you don't draw #1 the first time, then there are only 24 balls left, so p2 = 1/24, and p3 = 1/23. We need to multiply the q's (probabilities against) then subtract that product from 100%.So the probability of drawing #1 on any of the next 3 trials, starting with 25 balls is: 1 - ((1 - 1/25)(1 - 1/24)(1 - 1/23))= 1 - ((24/25)*(23/24)*(22/23)) = 12%. Notice the way I wrote it, the fractions cancel. Here's the interesting thing:The probability of getting #1 on the first try is 1/25 = 4%. The first or second try is 1 - ((24/25)*(23/24)) = 2/25 = 8%. Each trial will become an increased number over 25: 3/25 = 12%, 4/25 = 16%, etc. I'm not sure how you would generalize the formula though.
The probability of at least one event occurring out of several events is equal to one minus the probability of none of the events occurring. This is a binomial probability problem. Go to any binomial probability table with p=0.2, n=3 and the probability of 0 is 0.512. Therefore, 1-0.512 is 0.488 which is the probability of at least 1 sale.
An impossible event, with probability 0.
If an event keeps on occurring that means the event is at least annual. If it occurs more than once a year that makes the event perennial or biannual if it only occurs twice.
There is at least one video of such an event occurring.
The probability of an event is one when the event is absolutely certain to happen. It is more useful to understand why the sum of probabilities of an event adds up to one. If you have an event with a probability of p, then the probability of the event not occurring is 1 - p. This is because p plus (1 - p) is 1. That seems trivial, but consider a realistic example... Lets say you are subject to random drug testing, and that you are in a 100% pool such as might be mandated by 10CFR26, "Fitness for Duty in a Commericial Nuclear Power Plant". What is the probability that you will be chosen at least once in a year? Well, if sampling occurs five days a week, then there are 260 samples per year. The probability of being chosen in one sample is 1 in 260, or 0.003846. That means the probability of not being chosen is 259 in 260, or 0.9962. The probability, then, of not being chosen in one year is (259 in 260) raised to the 260th power, or 0.3672, which is about 95 in 260. Invert once again, and you see that the probability of being chosen at least one time in a year is 165 in 260, or 0.6328.
If p is the probability that an event will happen once, then the probability that it will happen just twice is p2. The probability it will happen 3 times is p3. The probability it will happen at least once ( ie once or twice or three times ore more times is p + p2 + p3 + ... = p(1-p). For "or" you add probabilities, for "and" you multiply probabilities.
If at least three flips are tails, there are two scenarios where we can obtain exactly four tails in five flips. Either the first four flips are tails and the last flip is heads, or the first flip is heads and the next four flips are tails. Each scenario has a probability of 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32. Therefore, the probability of obtaining exactly four tails in five flips if at least three are tails is 1/32 + 1/32 = 1/16.
The probability of landing on heads at least once is 1 - (1/2)100 = 1 - 7.9*10-31 which is extremely close to 1: that is, the event is virtually a certainty.
The probability level for an outcome is the probability that the outcome was at least as extreme as the one that was observed.
typically the carrier frequency has to be at least double the signal frequency but in order to get better results you want to choose a frequency that is at least 5 times the highest frequency you are designing for.
the electron cloud is least dense where the probability of finding an electron is LOWEST
Depends on the probability of reading any.