Binomial probability

From Wikipedia, the free encyclopedia

Binomial probability typically deals with the probability of several successive decisions, each of which has two possible outcomes.

1 Definition
2 Calculation
- 2.1 General equation
- 2.2 Example
3 Estimation
- 3.1 Binomial approximation
- 3.2 Poisson probability function
4 Connection to binomial theorem
5 See also
6 External links

[edit] Definition

The probability of an event can be expressed as a binomial probability if its outcomes can be broken down into two probabilities p and q, where p and q are complementary (i.e. p + q = 1) For example, tossing a coin can be either heads or tails, each which have a (theoretical) probability of 0.5. Rolling a four on a six sided die can be expressed as the probability (1/6) of getting a 4 or the probability (5/6) of rolling something else.

[edit] Calculation

If an event has a probability, p, of happening, then the probability of it happening twice is p², and in general pⁿ for n successive trials. If we want to know the probability of rolling a die and getting two fours and one other number (in that specific order) it becomes:

$\begin{matrix}P(\mbox{2 rolls of four and 1 other}) & = & P(\mbox{2 rolls of four}) P(\mbox{1 other}) \\ \ & = & P(\mbox{rolls of four})^2 P(\mbox{other})^1 \\ \ & = & (\frac{1}{6})^2 (\frac{5}{6})^1 \\ \ & = & 2.6% \end{matrix}$

However this is only sufficent for problems where the order is specific. If order is not important in the above example, then there are 3 ways that 2 rolls of four and 1 other could occur:

110
101
011

Where 1 represents a roll of four and 0 represents a non-four roll. Since there are 3 ways of achieving the same goal, so the probability is 3 times that of before, or 7.8%. If order doesn't matter, then there are ${n \choose r}$ (n choose r) possible configurations. This yields the general equation for binomial trials:

[edit] General equation

$P(\mbox{k success out of n trials}) = {n \choose k}p^kq^{n-k}$

Where p is the probability of a success, and q is the probability of a failure (which is complimentary to p, that is q=1-p.) This is simply a rewording of the binomial theorem.

[edit] Example

On a ten question multiple choice test, with 4 options per question, the probability of getting 5 and only 5 answers correct if the answers are guessed can be calculated like so:

p = 0.25

q = 1 - p = 0.75

k = 5

n = 10

$\begin{matrix}P(pass) & = & {n \choose k}p^kq^{n-k} \\ \ & = & {10 \choose 5}(0.25)^5(0.75)^5 \\ \ & \approx & 5.8%\end{matrix}$

Therefore if somebody guesses 10 answers on a multiple choice test with 4 options, they have about a 5.8% chance of getting 5 and only 5 correct answers. If 5 or more correct answers are needed to pass, then the probability of passing can be calculated by adding the probability of getting 5 (and only 5) answers correct, 6 (and only 6) answers correct, and so on up to 10 answers correct. This total probability is approximately 7.8%.

[edit] Estimation

There are various methods at estimating the binomial probability if the exponents are too large to calculate

[edit] Binomial approximation

Main article: Binomial distribution

One method is by approximating the probability to a binomial distribution. The requirements are that $np \ge 5$ and $nq \ge 5$ for an accurate answer. Approximation is done with the following equation:

$z \approx \frac{(k-\mu) \pm 0.5}{\sigma}$

Where $μ = n p$ and $\sigma = \sqrt{npq}$ (the standard deviation of the binomial approximation) and z is the corresponding z-score

[edit] Poisson probability function

Main article: Poisson distribution

Another possible method is approximating to a poisson distribution. The requirements are that $n \ge 150$ and that np and npq are within 10% of eachother. The formula is

$\Pr(n \mbox{ out of }k)\approx\frac{e^{-\lambda} \lambda^k}{k!},\,\!$

where $λ = n p$ .

[edit] Connection to binomial theorem

The equation for binomial probability is the same as the binomial theorem equation, which can be used to calculate terms in Pascal's triangle and the expansion of binomial equations of the form $(a + b) n$ . So, if the binomial is expanded for n = 2, we get

$\begin{matrix}(a+b)^2 & = & {2 \choose 0}a^2b^0 + {2 \choose 1}a^1b^1 + {2 \choose 2}a^0b^2 \\ \ & = & a^2 + 2ab + b^2 \end{matrix}$

Rewriting the equation in a trivial way:

1 a a + 2 a b + 1 b b

If a represents heads and b tails, then the above shows all the possibilities and the number of possible combinations. That is, there is one way to get two heads (aa), two ways to get a head and a tail (2ab) and one way to get two tails (bb) This applies for any degree of n. Since the sum of the coefficents in the equation (a₀ + a₁ + a₂) is the total possibilities, and since each unique case has the same probability, the probability of getting 1 occurrence of a and 1 of b (1 head and 1 tail on a coin) is 2 out of 4, or 0.5. The sum of coefficents for any binomial is 2^n − 1.