The Binomial Distribution ► Arrangements ► Remember the binomial theorem?

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

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

Example – using a calculator How many ways are there of arranging these? A A A B B B B B B n = 9r = 3 To calculate this, type “9” followed by “ n C r ” 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-defect  (1-p)(1-p)(1-p)  (1-p) 3

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

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

Binomial distribution ► We want P(X=3):  Can happen one way: 111  ppp p3p3 1 - good bulb 0 - defect bulb

Binomial distribution P(X=0): (1-p) 3 P(X=3): p 3 P(X=1):3p(1-p) 2 P(X=2): 3p 2 (1-p) 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  The probability of a good bulb is 0.7  What is the probability of there being 8 good bulbs in the test? n = 10 p = 0.7 r = 8

Binomial distribution - example n = 10p = 0.9r = 8 = (3 d.p.)

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

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?  The mean (expected value)? = (3 d.p.) The mean value of X is given by np = 10 x 0.7 = 7