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The answer to the first question is 2 / 3 , as is shown correctly by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1 / 2 .
In this situation, the event A can be analyzed by a conditional probability with respect to B. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B) [2] or occasionally P B (A).
The essay includes an example of a man trying to guess the ratio of "blanks" and "prizes" at a lottery. So far the man has watched the lottery draw ten blanks and one prize. Given these data, Bayes showed in detail how to compute the probability that the ratio of blanks to prizes is between 9:1 and 11:1 (the probability is low - about 7.7%).
Conditional probabilities, conditional expectations, and conditional probability distributions are treated on three levels: discrete probabilities, probability density functions, and measure theory. Conditioning leads to a non-random result if the condition is completely specified; otherwise, if the condition is left random, the result of ...
Given , the Radon-Nikodym theorem implies that there is [3] a -measurable random variable ():, called the conditional probability, such that () = for every , and such a random variable is uniquely defined up to sets of probability zero. A conditional probability is called regular if () is a probability measure on (,) for all a.e.
Each scenario has a 1 / 6 probability. The original three prisoners problem can be seen in this light: The warden in that problem still has these six cases, each with a 1 / 6 probability of occurring. However, the warden in the original case cannot reveal the fate of a pardoned prisoner.
The conditional probability the car is behind Door 2 given the player has picked Door 1 and has seen the host open Door 3 is defined as the probability the car is behind Door 2 in the cases the host opens Door 3 (1/3) divided by the probability of all cases where the host opens Door 3 (1/2), therefore the conditional probability is (1/3)/(1/2)=2/3.
This rule allows one to express a joint probability in terms of only conditional probabilities. [4] The rule is notably used in the context of discrete stochastic processes and in applications, e.g. the study of Bayesian networks, which describe a probability distribution in terms of conditional probabilities.