Sentences with phrase «probability equal outcomes»
It is a perfect resource for Probability equal outcomes.
https://www.tes.com/ Not exact matches
If you want a laugh equal to your fear check out present day dollars return for your portfolio with an upside probability equal to the negative outcome you are offsetting; it is crazy large.
The probability of winning a roll of the dice, for example, is equal to the proportion of winning outcomes relative to all possible ones.
The outcome of a given division is unpredictable, but in homeostasis the probabilities of producing two progenitor and two differentiating daughters are the same, so that on average, equal numbers of progenitors and differentiating cells are produced across whole population of progenitors.
As journalist Louis Menand put it, «The experts performed worse than they would have if they had simply assigned an equal probability to all three outcomes... Human beings who spend their lives studying the state of the world, in other words, are poorer forecasters than dart - throwing monkeys.»
Under Cardano's theory of fair gambling devices, equal numerical values are assigned to the probabilities of the ways in which an outcome can occur in a game of chance.
That is, the kurtosis of the model output would be flat, meaning pretty much equal probability of any outcome.
Should one ignore the very high probability of «equal = no - change» or «better than» outcomes?
The Wikipedia definition of the likelihood function is reasonable: the likelihood of a set of parameter values given some observed outcomes is equal to the probability [density] of those observed outcomes given those parameter values.
If Monte Carlo analysis were done on the climate models, I am sure that what we would see is pretty much equal probability for any outcome — ie flat kurtosis.
If we assume two complementary hypotheses H0 and H1, an experimental outcome O, know P (O H0) and P (O H1), and have an assumed prior probability ratio P (H0) / P (H1), we can calculate the posterior probability as follows: P (H0 O) / P (H1 O) = (P (O H0) / P (O H1)-RRB-(P (H0) / P (H1)-RRB- Take logs to convert that multiplication to an addition log (P (H0 O) / P (H1 O)-RRB- = log (P (O H0) / P (O H1)-RRB- + log (P (H0) / P (H1)-RRB- and we interpret this as the confidence in H0 over H1 after the observation is equal to the evidence inherent in the result of the experiment plus our confidence before the observation.