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