Lecture 18 A

GEM2900 Understanding Uncertainty and Statistical ThinkingNotes for lecture 18

Derivation of binomial probability function:

Recall the notation (n

x

)=

n!

x!(n− x)!

for the number of ways of choosing x objects from n without regard to order.Here i! is the product of integers up to i. Sometimes the notation nCx isalso used.

The binomial probability function is

P (X = x) =

(n

x

)px(1 − p)n−x x = 0, ..., n.

The probability of x successes and n − x failures in a particular order ispx(1 − p)n−x (probability p for each of the x successes, 1 − p for each of then− x failures, and multiply the probabilities because of independence). Weneed to multiply this expression by the number of ways of having x successes(i.e.

(nx

)) to get P (X = x).

Example: n = 3, x = 1. Consider the particular sequence of successes andfailures SFF where the notation means success on the first trial, failure onthe second and third trials. Clearly

P (SFF ) = P (S)P (F )P (F ) = p(1 − p)2

illustrating our rule of “probability of x successes and n − x failures in aparticular order is px(1 − p)n−x”. Now, how many ways can we have onesuccess in three trials? Clearly the success can be on the first trial, or thesecond, or the third. The possibilities are SFF , FSF and FFS. There are(nx

)=

(31

)= 3 possibilities. Multiplying p(1 − p)2 by 3 gives the probability

of one success.

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