Binomial DistributionBinomial Distribution

What the binomial distribution isWhat the binomial distribution is

How to recognise situations where the How to recognise situations where the binomial distribution appliesbinomial distribution applies

How to find probabilities for a given How to find probabilities for a given binomial distribution, by calculation and binomial distribution, by calculation and from tablesfrom tables

When to use the When to use the binomial distributionbinomial distribution Independent variablesIndependent variables

Pascal’s Triangle Pascal’s Triangle (a+b)(a+b)nn

1 11 1

1 2 11 2 1

1 3 3 11 3 3 1

1 4 6 4 11 4 6 4 1

1 5 10 10 5 11 5 10 10 5 1

10 ways to get to the 3rd position numbering each of the terms from 0 to 5. this can also be calculated by using nCr button on your calculator 5C2=10

nCrnCr

55CC00 11

55CC11 55

55CC22 1010

55CC33 1010

55CC44 55

55CC55 11

Pascal’s Triangle Pascal’s Triangle (a+b)(a+b)nn

1 11 1

1 2 11 2 1

1 3 3 11 3 3 1

1 4 6 4 11 4 6 4 1

1 5 10 10 5 11 5 10 10 5 1

nCrnCr n!n!÷(c!x(n-÷(c!x(n-c)!)c)!)

55CC00 5!5!÷(0!x5!)÷(0!x5!) 11

55CC11 5!5!÷(1!x4!)÷(1!x4!) 55

55CC22 5!5!÷(2!x3!)÷(2!x3!) 1010

55CC33 5!5!÷(3!x2!)÷(3!x2!) 1010

55CC44 5!5!÷(4!x1!)÷(4!x1!) 55

55CC55 5!5!÷(5!x0!)÷(5!x0!) 11

A coin is tossed 7 times. Find A coin is tossed 7 times. Find the probability of getting the probability of getting exactly 3 heads.exactly 3 heads.

We could do Pascal's triangle or we could calculate:We could do Pascal's triangle or we could calculate: 77CC3 3 x (P(H))x (P(H))77

The probability of getting a head is ½The probability of getting a head is ½

r

nnCr

3

737C 27.0

128

35

2

135

2

1

3

77

7

TASK TASK

Exercise A Page 61Exercise A Page 61

Unequal ProbabilitiesUnequal Probabilities A dice is rolled 5 times A dice is rolled 5 times What is the probability it will show What is the probability it will show

6 exactly 3 times?6 exactly 3 times?P(6’)=5/6P(6’)=5/6

P(6)=1/6 P(6)=1/6

103

535

C

rolls) 5in sixes 3(6

5

6

1

3

5 23

P

Task / HomeworkTask / Homework

Exercise B Page 62Exercise B Page 62

The Binomial distribution is all The Binomial distribution is all about success and failure. about success and failure.

When to use the Binomial DistributionWhen to use the Binomial Distribution

– A fixed number of trialsA fixed number of trials– Only two outcomes Only two outcomes

– (true, false; heads tails; girl,boy; six, not six …..)(true, false; heads tails; girl,boy; six, not six …..)

– Each trial is independentEach trial is independent

IF the random variable X has Binomial IF the random variable X has Binomial distribution, then we write X distribution, then we write X ̴ B(n,p)̴ B(n,p)

X

Sometimes you have Sometimes you have to use the Binomial to use the Binomial FormulaFormula

pqwhere

qpx

nxXP xnx

1

,)( )(

Eggs are packed in boxes of 12. The Eggs are packed in boxes of 12. The probability that each egg is broken is 0.35probability that each egg is broken is 0.35

Find the probability in a random box of Find the probability in a random box of eggs:eggs:there are 4 broken eggsthere are 4 broken eggs

figurest significan 3 to235.0

65.035.049565.035.04

12)4( 84)412(4

XP

Task / homeworkTask / homework

Exercise C Page 65Exercise C Page 65

Eggs are packed in boxes of 12. The Eggs are packed in boxes of 12. The probability that each egg is broken is 0.35probability that each egg is broken is 0.35Find the probability in a random box of Find the probability in a random box of eggs:eggs:

There are less than 3 broken eggsThere are less than 3 broken eggs

0151.001346.01225.06665.035.012005688.011

65.035.02

1265.035.0

1

1265.035.0

0

12

)2()1()0()3(

111

)10(2)11(1)12(0

XPXPXPXP

USING TABLES of the USING TABLES of the Binomial distributionBinomial distribution

An easier way to add up binomial An easier way to add up binomial probabilities is to use the probabilities is to use the cumulative binomial tablescumulative binomial tables

Find the probability of getting 3 successes in 6 trials,

when n=6 and p=0.3

n=6n=6 xx 00 11 22 33 44 55 66

P=0.3P=0.3 P(X=xP(X=x))

0.1170.11766

0.4200.42022

0.7440.74433

0.9290.92955

0.9890.98911

0.9990.99933

1.0001.000

n=6n=6 xx 00 11 22 33 44 55 66

P=0.3P=0.3 P(X=xP(X=x))

0.1170.11766

0.4200.42022

0.7440.74433

0.9290.92955

0.9890.98911

0.9990.99933

1.0001.000

http://assets.cambridge.org/97805216/05397/excerpt/9780521605397_excerpt.pdf

The probability of getting 3 or fewer successes is found by adding:P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1176 + 0.3026 + 0.3241 + 0.1852 = 0.9295

The probability of getting 3 or fewer successes is found by adding:P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1176 + 0.3026 + 0.3241 + 0.1852 = 0.9295

This is a cumulative probability.

Task / homeworkTask / homework

Exercise D page 67Exercise D page 67

Mean Mean variance and standard variance and standard

deviationdeviation μ μ = = ΣΣxx xx P(X= P(X=xx)=mean)=mean

This is the description of how to get the This is the description of how to get the mean of a discrete and random variable mean of a discrete and random variable defined in previous chapter.defined in previous chapter.

The mean of a random variable The mean of a random variable whos distribution is B(n,p) is whos distribution is B(n,p) is given as:given as:

μ μ =np=np

Mean, Mean, variancevariance & & standard standard deviationdeviation

σσ²=²=ΣΣxx² x P(X=² x P(X=xx) - ) - μμ² ² is the definition of variance, from the is the definition of variance, from the

last chapter of a discrete random last chapter of a discrete random variable.variable.

The variance of a random variable The variance of a random variable whose distribution is B(n,p)whose distribution is B(n,p)

σσ²=²= np(1-p) np(1-p) σσ==

)1( pnp

TASK / HOMEWORK TASK / HOMEWORK

Exercise EExercise E Mixed QuestionsMixed Questions Test Your selfTest Your self