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In this basic urn model in probability theory, the urn contains x white and y black balls, well-mixed together. One ball is drawn randomly from the urn and its color observed; it is then placed back in the urn (or not), and the selection process is repeated. [3] Possible questions that can be answered in this model are:
A read-once branching program can be represented by a multilinear polynomial which computes (over any field) on {0,1}-inputs the same Boolean function as the branching program, and two branching programs compute the same function if and only if the corresponding polynomials are equal. Thus, identity of Boolean functions computed by read-once ...
At each step, one ball is drawn uniformly at random from the urn, and its color observed; it is then returned in the urn, and an additional ball of the same color is added to the urn. If by random chance, more black balls are drawn than white balls in the initial few draws, it would make it more likely for more black balls to be drawn later.
This means that the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances (i.e., the square of the standard deviation is the sum of the squares of the standard deviations). [1]
Sum of the distance between the vertices and the difference of their colors is greater than k + 1, where k is a positive integer. Rank coloring If two vertices have the same color i, then every path between them contain a vertex with color greater than i Subcoloring An improper vertex coloring where every color class induces a union of cliques
The assignment problem consists of finding, in a weighted bipartite graph, a matching of maximum size, in which the sum of weights of the edges is minimum. If the numbers of agents and tasks are equal, then the problem is called balanced assignment, and the graph-theoretic version is called minimum-cost perfect matching.
The first column sum is the probability that x =0 and y equals any of the values it can have – that is, the column sum 6/9 is the marginal probability that x=0. If we want to find the probability that y=0 given that x=0, we compute the fraction of the probabilities in the x=0 column that have the value y=0, which is 4/9 ÷ 6/9 = 4/6. Likewise ...
The probability of the event that the sum + is five is , since four of the thirty-six equally likely pairs of outcomes sum to five. If the sample space was all of the possible sums obtained from rolling two six-sided dice, the above formula can still be applied because the dice rolls are fair, but the number of outcomes in a given event will vary.