Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
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第 1 到 3 筆結果,共 5 筆
第 26 頁
... passengers selects at random the carriage in which he will ride . What is the probability that there will be at least one passenger in each carriage ? b . Under the hypotheses of part a , what is the probability that exactly r of the ...
... passengers selects at random the carriage in which he will ride . What is the probability that there will be at least one passenger in each carriage ? b . Under the hypotheses of part a , what is the probability that exactly r of the ...
第 162 頁
... passengers choosing at random ( independently of the others ) one of the m carriages of the train . One passenger has m possibilities for the choice of his carriage , two passengers have m2 possibilities , ... , and p passengers have mo ...
... passengers choosing at random ( independently of the others ) one of the m carriages of the train . One passenger has m possibilities for the choice of his carriage , two passengers have m2 possibilities , ... , and p passengers have mo ...
第 165 頁
... passengers than carriages , then the probability that every carriage will be occupied is 0. Therefore the solution to part a must vanish when p < m , that is m3 - Since ( m ) 19 ― ' ) ( m − 1 ) 2 + ( m ) ( m ― 2 ) " - - ·· + ( − 1 ) m ...
... passengers than carriages , then the probability that every carriage will be occupied is 0. Therefore the solution to part a must vanish when p < m , that is m3 - Since ( m ) 19 ― ' ) ( m − 1 ) 2 + ( m ) ( m ― 2 ) " - - ·· + ( − 1 ) m ...
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A₁ A₂ An+m arrangements b₁ B₂ binomial coefficients binomial theorem bishops black squares C₁ chessboard chord circle coefficient color column compute the number Consequently consider corresponding customers denote determine the number diagonals digits dihedral angle divided divisible draw equally likely possible equation equidistant equivalence classes exactly example experiment favorable outcomes follows formula given Hence inclusion and exclusion intersection k-gons knights length mathematical induction maximum number multiple n-gon number of different number of favorable number of paths number of shortest obtain pairs partition passengers plane polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks S₁ segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose T₂ total number triangle unfavorable values vertex vertices