By Vladimir I. Rotar

Uncomplicated NotionsSample area and EventsProbabilitiesCounting TechniquesIndependence and Conditional ProbabilityIndependenceConditioningThe Borel-Cantelli TheoremDiscrete Random VariablesRandom Variables and VectorsExpected ValueVariance and different Moments. Inequalities for DeviationsSome simple DistributionsConvergence of Random Variables. The legislation of enormous NumbersConditional ExpectationGenerating features. BranchingRead more...

summary: simple NotionsSample house and EventsProbabilitiesCounting TechniquesIndependence and Conditional ProbabilityIndependenceConditioningThe Borel-Cantelli TheoremDiscrete Random VariablesRandom Variables and VectorsExpected ValueVariance and different Moments. Inequalities for DeviationsSome uncomplicated DistributionsConvergence of Random Variables. The legislations of huge NumbersConditional ExpectationGenerating services. Branching tactics. Random stroll RevisitedBranching strategies producing services Branching procedures Revisited extra on Random WalkMarkov ChainsDefinitions and Examples. chance Distributio

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**Example text**

1) P(Ac ) 1 − P(A) If the odds ratio is k, it is common to say that the “odds are k to 1” in favor of A. For instance, in the last example, for the switching strategy, the odds are 2 to 1 in favor of getting the car. , nk , respectively. So, n1 + n2 + ... + nk = n. Let Nn be the number of all possible divisions. We show that n! Nn = . n2 ! · · · nk ! Indeed, let us imagine that we are arranging all objects in a particular order, and we are doing it in the following way. First, we divide the objects into the groups mentioned in one of the Nn ways.

A point (a, b) is chosen from the square R = {0 ≤ a ≤ 1, 0 ≤ b ≤ 1}. Suppose that the probability distribution on R is uniform; that is, the probability that a point (a, b) comes from a region in R, equals the area of this region. (a) Find the probabilities that a ≥ b and 2a ≥ b. (b) Consider the quadratic equation x2 + 2ax + b = 0 with the coefficients (a, b). Find the probability that the equation has real solutions; has only one real solution. 11. Prove that P(ABc ) = P(A) − P(AB), and draw an illustration picture.

1a) We treat k draws as trials such that each may have one of n possible results (balls). In accordance with the basic counting principle, the total number of outcomes |Ω| = nk . 1b) The total number of outcomes is section. 2, we have shown that the number of all outcomes is (n)k = n(n − 1) · · · (n − k + 2)(n − k + 1). 1, the number of outcomes equals n k = n! (n−k)! Next, we show that the same model may be presented in terms of the occupancy problem that consists in distributing k objects (we will call them particles) into n groups; we will identify them with boxes.