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To get a probability of getting X heads in a sequence of N tosses, several things have to happen. If the probability of a head on a single trial is p and then the probability of a tail is q=(1-p) Then the result of X heads and N-X tails on any one trial has a probability calculated by finding the product of multiplying p times itself X times (p^X), and q times itself (N-X) times (q^(N-X)) giving (p^X q^(N-X)). However, there are many sequences which match this description. By the methods of CombinaTorics, we can find that there are N!/(X!*(N-X)!) different combinations with X heads and N-X tails. So, the probability of X heads is |
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To get a probability of getting X heads in a sequence of N tosses, several things have to happen. If the probability of a head on a single trial is p and then the probability of a tail is q=(1-p) Then the result of X heads and N-X tails on any one trial has a probability calculated by finding the product of multiplying p times itself X times (p^X), and q times itself (N-X) times (q^(N-X)), that is (p^X q^(N-X)). However, there are many sequences which match this description. By the methods of CombinaTorics, we can find that there are N!/(X!*(N-X)!) different combinations with X heads and N-X tails. So, the probability of X heads is |
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