BINOMIAL PROBABILITY DISTRIBUTION

Binomial Probability Distribution A binomial experiment is a probability experiment with the

following characteristics: – The experiment has n identical trials.– Two outcomes are possible on each trial – one trial is

termed a success and the other is termed a failure.– The probability of a success occurring on each trial is

p. This probability p is the same on each trial.– Since the outcome must either be a success or

failure, a failure is the complement of a success and the probability of a failure is 1-p. (Some texts refer to this probability as q, that is, q=1-p).

– The trials are independent of each other.

Given the above conditions:• The binomial probability distribution provides the probability

of x successes in n trials, where x=0, 1 ,2, 3, … , n.• Note that there are only two parameters that determine

binomial probabilities:n = the number of trials. p = the probability of success.

• Successive trials must be independent of each other. That is, the outcome of any one trial must not affect the probability of success or failure for any other trial. P (success failure on any other trial) = pP (success success on any other trial) = p

i

x f(x)0 1/8 = 0.125

1 3/8 = 0.375

2 3/8 = 0.375

3 1/8 = 0.125

Total 8/8 = 1.000

Example – number of females selected in a random sample of size 3 from a large population of half males and half females.

The above distribution is a binomial probability distribution with success defined as selecting a female. There are n = 3 independent trials, the probability of success is p = 0.5, and x is the number of successes. In this experiment, selecting a male is termed a failure, and the probability of selecting a male is 1-p = 1-0.5 = 0.5.

x is the number of females selected and f(x) is the probability of x females being selected

Formula for Binomial Probability

nCombinationCx

nnnnwhere

ppnCxxf

or

ppxnx

nxf

xnx

xnx

1!0)1)(2)....(2)(1(!

:)1.(.)(

)1.(.)!(!

!)(

)(

)(

If n is the number of trials of the binomial experiment and p is the probability of success, then the probability of x successes in n trials of the experiment is given by the probability function f(x), defined as follows:

Using the binomial formula

125.0125.01)125.0)(1(1231

123)5.01(5.0)!03(!0

!3)0(

375.0125.03)5.0)(25.0(112123)5.01(5.0

)!23(!2!3)2(

)03(0

)23(2

f

f

Combinations and permutations

Permutations – the number of ways of arranging N objects, taken n at a time, where the order of the objects is taken into account, is:

Where is the number of possible combinations of N objects, taken n at a time, where the order of the objects does not matter.

)!(!!nN

NnCP N

n

N

n

CN

n

)!(!!nNn

NCN

n

Rationale for the binomial formulaProbability of x successes and (n-x) failures is

This is and represents the probability of any particular sequence of x successes and (n-x) failures.

And there are ways of arranging these x successes and (n-x) failures. To obtain the probability of x successes in n trials, multiply the probability of any particular sequence by this combination.

)1(...)1()1()1(... pppppppp n – x times x times

)1( pp xnx

Cn

x

Example – selection of Saskatchewan workers, classified by years of education and wages and salaries

From all these workers, randomly select 13 workers with 14-17 years of education. What is the probability that exactly 8 of these will have incomes of $45,000 or more? Probability of 8 or more?

A random sample from a large population means that successive selections are independent of each other. There are n = 13 workers selected. If success is defined as the probability of selecting a worker with an income of $45,000 or more, the probability of success p = 82/230 = 0.357.

Probability of 8 with $45,000 or more income = 0.0373. See the following slides for the calculation.

Using the formula

00000154.0)1)(00000153.0(!0!13

!13)357.01(357.0)!1313(!13

!13)13(

000035827.0)643.0)(000004286.0(1!12

!1213)357.01(357.0)!1213(!12

!13)12(

000387139.0)413449.0)(000012005.0(12!11

!111213)357.01(357.0)!1113(!11

!13)11(

002556743.0)265847707.0)(000033627.0(123!10

!10111213)357.01(357.0)!1013(!10

!13)10(

011512347.0)170940076.0)(000094192.0(1234!9

!910111213)357.01(357.0)!913(!9

!13)9(

037323.0)109914469.0)(000263843.0(12345!8

!8910111213)357.01(357.0)!813(!8

!13)8(

)1313(13

)1213(12

)1113(11

)1013(10

)913(9

)813(8

f

f

f

f

f

f

Using the formula

00000154.0)1)(00000153.0(!0!13

!13)357.01(357.0)!1313(!13

!13)13(

000035827.0)643.0)(000004286.0(1!12

!1213)357.01(357.0)!1213(!12

!13)12(

000387139.0)413449.0)(000012005.0(12!11

!111213)357.01(357.0)!1113(!11

!13)11(

002556743.0)265847707.0)(000033627.0(123!10

!10111213)357.01(357.0)!1013(!10

!13)10(

011512347.0)170940076.0)(000094192.0(1234!9

!910111213)357.01(357.0)!913(!9

!13)9(

037323.0)109914469.0)(000263843.0(12345!8

!8910111213)357.01(357.0)!813(!8

!13)8(

)1313(13

)1213(12

)1113(11

)1013(10

)913(9

)813(8

f

f

f

f

f

f

Probabilities to 4 decimal places Number of successes (x)

Probability of x or f(x)

8 0.0373

9 0.0115

10 0.0026

11 0.0004

12 0.0000

13 0.0000

The probability of 8 or more successes is the sum of the probabilities of 8, 9, 10, 11, 12, or 13 successes. This is 0.0373 + 0.0115 + 0.0026 + 0.0004 + 0.0000 + 0.0000 = 0.0518.

1. Roll a die 3 times. X=# of sixes. S=a six, N=not a sixNo six: (x=0)

NNN (5/6)(5/6)(5/6)One six: (x=1)

NNS (5/6)(5/6)(1/6) NSN same SNN same

Two sixes: (x=2)NSS (5/6)(1/6)(1/6)SNS sameSSN same

Three sixes: (x=3)SSS (1/6)(1/6)(1/6)

Other Examples

Binomial distribution• x f(x) 0 (5/6)3

1 3(1/6)(5/6)2

2 3(1/6)2(5/6) 3 (1/6)3

xx

xxf

3

65

613

)(

2. Toss a die 5 times. X=# of sixes.Find P(X=2)S=six N=not a sixSSNNN 1/6*1/6*5/6*5/6*5/6=(1/6)2(5/6)3

SNSNN 1/6*5/6*1/6*5/6*5/6=(1/6)2(5/6)3

SNNSN 1/6*5/6*5/6*1/6*5/6=(1/6)2(5/6)3

SNNNSNSSNN etc.NSNSNNSNNsNNSSNNNSNSNNNSS

2 31 5( 2) 10*6 6

P x

[P(S)]# of S

10 ways to choose 2 of 5 places for S. __ __ __ __ __

5 5! 5! 5*4*3! 102 2!(5 2)! 2!3! 2*1*3!

[1-P(S)]5 - # of S

3. Roll a die 20 times. X=# of 6’s,n=20, p=1/6

4. Flip a fair coin 10 times. X=# of heads

20

4 16

20 1 5( )6 6

20 1 5( 4)4 6 6

x x

f xx

p x

1010

2110

21

2110

)(

xxxf

xx

5. Pumpkin seeds germinate with probability 0.93. Plant n=50 seeds X= # of seeds germinating

xx

xxf

5007.093.0

50)(

248 07.093.04850

)48(

XP

Sometimes you have to use the Binomial Formula

pqwhere

qpxn

xXP xnx

1

,)( )(

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

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

figurest significan 3 to235.0

65.035.049565.035.04

12)4( 84)412(4

XP

6.

Population Parameters of a Binomial Distribution

Mean: E(x)= = np Expected value

Variance: 2 = npqStandard Deviation: = √npq

Finding Mean, Variance and Standard Deviation

7. In Pittsburgh, 57% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. What can you conclude?

Solution: There are 30 days in June. Using n=30, p = 0.57, and q = 0.43, you can find the mean variance and standard deviation as shown.

Mean: = np = 30(0.57) = 17.1Variance: 2 = npq = 30(0.57)(0.43) = 7.353Standard Deviation: = √npq = √7.353 ≈2.71