The Binomial DistributionArrangements 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+ bbbbArrangements of 4 AsArrangements of 3 As and 1 BArrangements of 2 As and 2 BsArrangements of 1 A and 3 BsArrangements of 4 BsThe 2nd line contains terms corresponding to a3b so coefficient isThe 3rd line contains terms corresponding to a2b2 so coefficient isThe 4th line contains terms corresponding to ab3 so coefficient isThe 1st line contains terms corresponding to a4 so coefficient isThe 5th line contains terms corresponding to b4 so coefficient is

ArrangementsHow many ways are there of arranging these?A A A B B B B B Bn = 9 r = 3

Example using a calculatorHow many ways are there of arranging these?A A A B B B B B Bn = 9 r = 3 To calculate this, type 9 followed by nCr followed by 3 and press equals?Use your calculator to work outExplain your answer.

Binomial TheoremA general rule for any expansion isA special case occurs when a=1 and b=x

The Binomial DistributionThe binomial distribution is a discrete distribution defined by 2 parametersthe number of trials - nthe probability of a success - pWRITTEN : which means the discrete random variable X is binomially distributed

Binomial distribution - exampleTesting for defects with replacementHave many light bulbsPick one at random, test for defect, put it backPick one at random, test for defect, put it backIf there are many light bulbs, do not have to replace

Binomial distributionConsider 3 trials n=3p is the probability of picking a good bulbso (1-p) is the probability of picking a defect bulbthe random variable X is the measure of the number of good bulbsIf we want P(X=0):Can happen one way: defect-defect-defect(1-p)(1-p)(1-p)(1-p)3

Binomial distributionIf we want P(X=1):Can happen three ways: 100, 010, 001p(1-p)(1-p)+(1-p)p(1-p)+(1-p)(1-p)p3p(1-p)2

Binomial distributionIf we want P(X=2):Can happen three ways: 110, 011, 101pp(1-p)+(1-p)pp+p(1-p)p3p2(1-p)1 - good bulb0 - defect bulb

Binomial distributionWe want P(X=3):Can happen one way: 111pppp31 - good bulb0 - defect bulb

Binomial distributionP(X=0): (1-p)3P(X=3): p3P(X=1):3p(1-p)2P(X=2): 3p2(1-p)r is the number of good bulbs

Binomial distribution functionIn general, the binomial distribution function is given by:

Binomial distribution - exampleTesting for defects with replacementSuppose 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 = 10p = 0.7r = 8

Binomial distribution - examplen = 10p = 0.9r = 8= 0.233 (3 d.p.)

Binomial distributionTypical shape of binomial:SymmetricMean and Mode approx = p*nrP

Binomial distribution - expected valueFor the binomial distribution

The mean value of X is given by np

this is also the expected value of X - E(X)

Binomial distribution - exampleTesting for defects with replacementSuppose 10 bulbs were testedThe probability of a good bulb is 0.7What is the probability of there being 8 good bulbs in the test?

The mean (expected value)?= 0.233 (3 d.p.)The mean value of X is given by np= 10 x 0.7 = 7