yes
Queueing Theory Calculator is a simple, yet powerful tool to process queueing models calculations, Erlang formulas for queues.
good question.
plese send me commenication si model papers
The time taken between your posting this question and receiving a reply.
In theoretical probability, the probability is determined by an assumed model (for example, the normal distribution). (compare with empirical probability)
No.
Provided that the correct model is used, the theoretical probability is correct. The experimental probability tends towards the theoretical value as the number of trials increases.Provided that the correct model is used, the theoretical probability is correct. The experimental probability tends towards the theoretical value as the number of trials increases.Provided that the correct model is used, the theoretical probability is correct. The experimental probability tends towards the theoretical value as the number of trials increases.Provided that the correct model is used, the theoretical probability is correct. The experimental probability tends towards the theoretical value as the number of trials increases.
In almost all probability it will NOT fit. But you haven't mentioned the Inspiron model number in your question.
As the number of times that the experiment is conducted increases, the experimental probability will near the theoretical probability - unless there is a problem with the theoretical model.
it is basically a model that shows the probability that an eletron exists in that position
To find the probability of getting at least 6 correct answers on a 10-question multiple-choice exam where each question has 5 choices (with only one correct answer), we can model this situation using the binomial probability formula. The probability of guessing correctly on each question is ( p = \frac{1}{5} ) and incorrectly is ( q = \frac{4}{5} ). We need to calculate the sum of probabilities for getting exactly 6, 7, 8, 9, and 10 correct answers. Using the binomial formula, the probability ( P(X = k) ) for each ( k ) can be computed, and then summed to find ( P(X \geq 6) ). The resulting probability is approximately 0.0163, or 1.63%.
ou