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Topic: Problem on probability theory

All greetings! The problem from last year's entrance examination in  interested: are Abreast allocated m subjects. Are casually selected k subjects, k <m. The Random variable X is equal to an amount of such subjects i that i is selected, and all its neighbors are not selected. To find mathematical expectation X. I twist some days, there should be rather simple decision, but I it do not see. Tried to build allocation X explicitly, but for this purpose it is necessary to find number of methods to select k subjects so that exactly X from the selected had no selected neighbors. For this amount of methods the difficult recursion is visible only, with explicit expression there are problems. Counted  for different k and m statistically, saw that dependence from k well lays down on a cubic parabola... Found it on several points in a general view. But it did not approach me to the analytical decision. If who has ideas, share!

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Re: Problem on probability theory

Hello, kfmn, you wrote: K> All greetings! K> the problem from last year's entrance examination in  interested: K> are Abreast allocated m subjects. Are casually selected k subjects, k <m. The Random variable X is equal to an amount of such subjects i that i is selected, and all its neighbors are not selected. To find X. K> I Twist some days, there should be rather simple decision, but I it do not see. K> tried to build allocation X explicitly, but for this purpose it is necessary to find number of methods to select k subjects so that exactly X from the selected had no selected neighbors. For this amount of methods the difficult recursion is visible only, with explicit expression there are problems. Counted  for different k and m statistically, saw that dependence from k well lays down on a cubic parabola... Found it on several points in a general view. K> but it did not approach me to the analytical decision. K> if who has ideas, share! Like as it is not so difficult. Though, can, I and was mistaken. In total  at us a C (m, k). . probability of obtaining what or samplings are p = 1/C (m, k) Matozhidanie  are the total on all samplings A of values p*X (A). Where X (A) there is a value X on sampling X. Everyone X (A) in turn can be considered as the total on all positions i from 1 to m. Thus i- the item will be or 0 or p. p will be if the position i is selected from sampling A and adjacent positions are not present. Then Matozhidanie  is the total of the totals. At first the total on samplings A, and then the total on positions i. We make here swap of the totals. Will add at first on positions i, and then on samplings A. And here it becomes already simple. For a position 1 will be not 0, and p only for those  A from which the position 2 is not selected. Such  A will be a C (m-2, k-1) pieces. . for a position 1 total item will be p*C (m-2, k-1) / will be similar for a position m. And here for average positions (. for 2. ., m-1) into the total enter p only for  where there are no adjacent elements, which already 2. Such  A will be a C (m-3, k-1) pieces. Total the total sum is p * (2*C (m-2, k-1) + (m-2) * a C (m-3, k-1)) the Formula like the simple.

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Re: Problem on probability theory

Hello, baily, you wrote: B> Like as it is not so difficult. Though, can, I and was mistaken. B> all  at us a C (m, k). . probability of obtaining what or samplings are p = 1/C (m, k) B> Matozhidanie  are the total on all samplings A of values p*X (A). Where X (A) is value X on X. B> X (A) in turn can be considered Everyone as the total on all positions i from 1 to m. Thus i- the item will be or 0 or p. B> p will be if the position i is selected from sampling A and adjacent positions are not present. B> Then Matozhidanie  is the total of the totals. At first the total on samplings A, and then the total on i. B> we Make here swap of the totals. Will add at first on positions i, and then on samplings A. B> And here it becomes already simple. For a position 1 will be not 0, and p only for those  A from which the position 2 is not selected. Such  A will be a C (m-2, k-1) pieces. B> . for a position 1 total item will be p*C (m-2, k-1) / B> will be similar for a position m. B> And here for average positions (. for 2..., m-1) into the total enter p only for  where there are no adjacent elements, which already 2. Such  A will be a C (m-3, k-1) pieces. B> total the total sum is p * (2*C (m-2, k-1) + (m-2) * a C (m-3, k-1)) B> the Formula like the simple. The class, here with swap of the totals I also did not guess a trick! All tried for specific X to find number , and it was not necessary. Thanks!