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Let X be the random variable representing the number of successes in n binomial trials. The binomial probability P(X=x) can be calculated by using the following formula.
P(X=x)=_nC_xp^xq^(n-x)
In this formula, P(X=x) is the probability that the random variable X is equal to x, which means that there are exactly x successes. Additionally, p and q are the probabilities of success and failure, respectively, and _nC_x is the binomial coefficient.
Note that there are n independent trials, of which x are successes. This means that the difference between n and x gives the number of failures. rc Number of Succeses:& x Number of Failures:& n- x Furthermore, let p be the probability of success and q be the probability of failure in one individual trial. Therefore, by the Multiplication Rule of Probability and the fact that trials are independent, the probability of x successes is given by multiplying p by itself x times. Probability of xSuccesses: p* p* ... * p_(xtimes)=p^x Similarly, because there are n- x failures, the probability of n- x failures is given by multiplying q by itself n- x times. Probability of n- xFailures: q* q* ... * q_(n- xtimes)=q^(n- x) Therefore, the probability of getting exactly x successes and n- x failures is given by the product of p^x and q^(n- x) p^x q^(n- x) By the Commutative Property of Multiplication, any combination of p and q variables can be rearranged. Therefore, the expression is valid for any fixed sequence of x successes and n- x failures. However, note that this is the probability of only one of the possible sequences.
Consider an experiment that consists of four independent trials. The sequences in which two successes S and two failures F occur are as follows. {S S F F, S F S F, S F F S, F S S F, F S F S, F F S S} Note that as long as two successes occur, the order of the outcomes is not important. This can be simplified by considering how many ways the two successes can be arranged within the four trials since the remaining trials will automatically be failures. In such a situation, the combination formula can be used.
n= 4, r= 2
Subtract term
Write as a product
Cross out common factors
Cancel out common factors
Multiply
2!=2
Calculate quotient
This means that 6 different arrangements are possible, which corresponds to the number of sequences listed before. By following the same reasoning, different sequences of n trials with x successes can be found by calculating the number of combinations of x out of n. _nC_x=n!/x!( n- x)!
The probability of obtaining a fixed sequence of x successes out of n trials is given by the product of p to the power of x and q to the power of n- x. p^x q^(n- x) Moreover, there are _nC_x possible sequences with x successes. Therefore, by the Addition Rule of Probability, the probability P(X= x) of getting any of the possible sequences is given by adding the probability of getting one sequence _nC_x times. P(X=x)=p^x q^(n-x)+...+p^x q^(n-x)_(_nC_x times) ⇓ P(X=x)= _nC_x p^x q^(n-x)