A
statistical function to calculate the
probability of getting
n successes out of
N attempts in a
mutually exclusive,
dichotomous,
independent,
random function with the probability of
p.
Well, that's a load of gibberish, isn't it? Allow me to explain:
This can be described mathematically by the following function:
N!
P(n) = ---------- pn(1-p)N-n
n!(N - n)!
|_________|
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|
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| Side note: This is the binomial coefficient |
| of N and n, or "N choose n" |
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P(n) is the probability of achieving exactly n successes. p is the probability of success in one single case. N is the number of trials.
An example: If I flip a coin seven times, what is the probability that it will land heads up twice? This gives us the following values:
N = 7
n = 2
p = 0.5 (it's a 50/50 chance of heads/tails)
Insert these values into the function:
7!
P(2) = ---------- 0.52(1-0.5)7-2 = 0.1640625
2!(7 - 2)!
The probability is approximately 16%.
This function is often used to calculate the probability to achieve at least m successes in N tries. This can be done in the following way:
N
---
\ P(ni)
/
---
i=m
In the previous example, if I had wanted to calculate the probability to achieve at least 5 heads up, here's the formula:
7
---
\ P(ni) = P(5) + P(6) + P(7)
/
---
i=5
This will cause a distribution not entirely unlike the normal distribution (the gaussian distribution).