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.
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.
We need to determine the separate event. Let A = obtaining four tails in five flips of coin Let B = obtaining at least three tails in five flips of coin Apply Binomial Theorem for this problem, and we have: P(A | B) = P(A ∩ B) / P(B) P(A | B) means the probability of "given event B, or if event B occurs, then event A occurs." P(A ∩ B) means the probability in which both event B and event A occur at a same time. P(B) means the probability of event B occurs. Work out each term... P(B) = (5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0 It's obvious that P(A ∩ B) = (5 choose 4)(½)4(½) since A ∩ B represents events A and B occurring at the same time, so there must be four tails occurring in five flips of coin. Hence, you should get: P(A | B) = P(A ∩ B) / P(B) = ((5 choose 4)(½)4(½))/((5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0)
Depends on the probability of reading any.
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875
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.
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.
We need to determine the separate event. Let A = obtaining four tails in five flips of coin Let B = obtaining at least three tails in five flips of coin Apply Binomial Theorem for this problem, and we have: P(A | B) = P(A ∩ B) / P(B) P(A | B) means the probability of "given event B, or if event B occurs, then event A occurs." P(A ∩ B) means the probability in which both event B and event A occur at a same time. P(B) means the probability of event B occurs. Work out each term... P(B) = (5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0 It's obvious that P(A ∩ B) = (5 choose 4)(½)4(½) since A ∩ B represents events A and B occurring at the same time, so there must be four tails occurring in five flips of coin. Hence, you should get: P(A | B) = P(A ∩ B) / P(B) = ((5 choose 4)(½)4(½))/((5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0)
The probability level for an outcome is the probability that the outcome was at least as extreme as the one that was observed.
Depends on the probability of reading any.
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.
Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875