# BinomialDistribution

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Difference (from prior major revision) (author diff)

Changed: 3c3
 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
 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

The Binomial distribution can be described as the sum of a specific number of independent trials, each of which results in either a zero or a one with constant probability. It provides a reasonable description of coin tossing experiments, among others.

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

```   N!/(X!*(N-X)!) p^X q^(N-X)
```

To the author: First q=1-p, but more important your formula the probabilty of X heads out of N trials is WRONG. See BinomialDistribution/Revisited RoseParks

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