The Binomial Distribution Arrangements Remember the binomial theorem

The Binomial Distribution ►Arrangements ►Remember the binomial theorem?

Expanding using arrangements Arrangements of 4 As (a+b)4 = aaaa Arrangements of 3 As and 1 B + aaab + aaba + abaa + baaa Arrangements of 2 As and 2 Bs + aabb + abab + abba + baab + baba + bbaa + abbb + babb + bbab + bbba + bbbb Arrangements of 1 A and 3 Bs Arrangements of 4 Bs The 1 st line contains terms corresponding to a 4 so coefficient is The 2 nd line contains terms corresponding to a 3 b so coefficient is The 3 rd line contains terms corresponding to a 2 b 2 so coefficient is The 4 th line contains terms corresponding to ab 3 so coefficient is The 5 th line contains terms corresponding to b 4 so coefficient is

Arrangements AAABBBBBB How many ways are there of arranging these? n=9 r=3

Example – using a calculator AAABBBBBB How many ways are there of arranging these? n=9 r=3 To calculate this, type “ 9” followed by “n. Cr” followed by “ 3” and press equals? Use your calculator to work out Explain your answer.

Binomial Theorem A general rule for any expansion is… A special case occurs when a=1 and b=x …

The Binomial Distribution • The binomial distribution is a discrete distribution defined by 2 parameters • the number of trials - n • the probability of a success - p WRITTEN : … which means the discrete random variable X is binomially distributed

Binomial distribution - example ► Testing § § for defects “with replacement” Have many light bulbs Pick one at random, test for defect, put it back If there are many light bulbs, do not have to replace

Binomial distribution ► Consider 3 trials § n=3 ►p is the probability of picking a good bulb § so (1 -p) is the probability of picking a defect bulb ► the random variable X is the measure of the number of good bulbs ► If we want P(X=0): § Can happen one way: defect-defect § (1 -p)(1 -p) § (1 -p)3

Binomial distribution For simplicity: 1 - good bulb 0 - defect bulb ► If we want P(X=1): § Can happen three ways: 100, 010, 001 § p(1 -p)+(1 -p)p(1 -p)+(1 -p)p § 3 p(1 -p)2

Binomial distribution ► If we want P(X=2): 1 - good bulb 0 - defect bulb § Can happen three ways: 110, 011, 101 § pp(1 -p)+(1 -p)pp+p(1 -p)p § 3 p 2(1 -p)

Binomial distribution ► We want P(X=3): 1 - good bulb 0 - defect bulb § Can happen one way: 111 § ppp § p 3

Binomial distribution P(X=0): P(X=1): P(X=2): P(X=3): (1 -p)3 3 p(1 -p)2 3 p 2(1 -p) p 3 r is the number of good bulbs

Binomial distribution function § In general, the binomial distribution function is given by: or q = 1 -p

Binomial distribution - example ► Testing for defects “with replacement” § Suppose 10 bulbs were tested n = 10 § The probability of a good bulb is 0. 7 p = 0. 7 § What is the probability of there being 8 good bulbs in the test? r=8

Binomial distribution - example n = 10 p = 0. 9 r=8 = 0. 233 (3 d. p. )

Binomial distribution ► Typical shape of binomial: § Symmetric § Mean and Mode approx = p*n P r

Binomial distribution - expected value ► For the binomial distribution The mean value of X is given by np … this is also the expected value of X - E(X)

Binomial distribution - example ► Testing for defects “with replacement” § Suppose 10 bulbs were tested § The probability of a good bulb is 0. 7 § What is the probability of there being 8 good bulbs in the test? = 0. 233 (3 d. p. ) § The mean (expected value)? The mean value of X is given by = 10 x 0. 7 = 7 np