This document discusses binomial distribution concepts including: - Listing possible values of a discrete random variable - Calculating probabilities of events using the binomial probability formula - Determining the mean, variance, and standard deviation of a binomial distribution Several examples are provided to demonstrate calculating probabilities and key binomial distribution parameters like the number of successes, trials, probability of success, and applying the relevant formulas.

1. Cikgu Hanini Suhaila Hamsan Binomial Distribution puterizamrud@gmail.com
List of possible valuesof adiscrete random
variable
1 If x representthe numberof pupilsscoring
12A ina group of 5 pupils,listall the possible
values
5!
2 If the numberof timesof gettingthe number
1 whentossingafairdice three times, listall
possible values
3!
Determine the probabilityof aneventina
Binomial Distribution
1 Givenp=0.7 q = 0.3
N= 5 r= 3
FindP ( x=3)
= 5𝑐3 0.730.32
=0.3087
2 The ProbabilitythatAhmadwill be late forthe
meetingis0.7 Findthe probabilitythat
Ahmadwill be late for5 out of sevenmeeting
P= 0.7 q= 0.4
N=7 r= 5
= 7𝑐5 0.750.42
=
3 The ProbabilitythatSinYingscoring1 A for
Englishinthe monthlytestis0.4 . Findthe
probabilitythatSinYingwill be scoring1A for
Englishtwice out of 6 monthlytest
P= 0.4 q= 0.6 n= 6 r=2
=6𝑐2 0.420.64
= 15 (0.16)(0.1296)
=0.3110
4 The probabilitythatAhmadwearsbatikshirt
for a meetingis0.54 Findthe probabilitythat
Ahmadwearbatik 3 out of sevenmeeting
= 7𝑐3 0.5430.554
=
5 In SMK SimpangLimathe probabilityof pupil
beinglate is8% . Calculate the probability
that a group of 10 beinglate
a) Exactlytwo
P=0.08 n =10
P ( x=2) = 10𝑐2 0.0820.928
= 45( 0.0064) (0.513)
= 0.1478
b) Lessthan two
P ( x < 2) = 𝑃 ( 𝑥 = 0) + 𝑃 ( 𝑥 = 1)
=10𝑐0 0.0800.9210 + 10𝑐1
0.0810.929
=1 ( 1) 0.434 + 10(0.08)(0.472)
=0.434 + 0.378
=0.812
6 The Probabilitythat eachshotfiredbyRamli
hitsa target is1/3
a) If Ramli fires10 shotfindthe
probabilitythatexactly2 shots hitthe
target
P = 1/3 n=10
P ( x= 2) =10𝑐2 0.33320.6678
=0.1951Type equation here.
b) If Ramli firesn shotsthe probability all
the n shotshit the targetis 1/243
Findthe value of n
P =1/243 n = n
P ( x=n )
= 𝑛𝑐 𝑛
1
3
𝑛
(1 −
1
3
) 𝑛−𝑛 =
1
243
= (
1
3
) 𝑛 =
1
243

2. Cikgu Hanini Suhaila Hamsan Binomial Distribution puterizamrud@gmail.com
7
The resultof a studyshowthat 20%
of the pupil ina citycycle to school .
If 8 pupilsfromthe cityare chosenat
randomcalculate the probability
P= 0.2 n=8
a) Exactly2 of themcycle to school
P ( x=2) =8𝑐2 0.220.86
= 28 ( 0.04) (0.262144)= 0.2936
b) Lessthan 3 of themcycle
= P(x< 3)
= 𝑃( 𝑥 = 0) + 𝑃 ( 𝑥 = 1 )
+ 𝑃 ( 𝑥 = 2)
= 8𝑐0 0.2000.88 + 8𝑐1 0.210.87
+ 0.2936
= 1( 1) (0.167772) + 8( 0.2)
(0.2097) + 0.2936
= 0.1678 + 0.3355 + 0.2936
= 0.7969
3−𝑛= 243−1=35(−1)3−5
𝑛 = 5
9 It isknownthat 18 out of 30 studentsina
classlike toread duringtheirfree time
9 studentsare selectedatrandomfromthe
class. Findthe probabilitythat
a) All the selectedstudentslike toread
duringtheirfree time
P ( x=9)
= 9𝑐9 0.6 90.40
= 1 ( 0.01008)(1)
= 0.01008
b) At least7 of the selectedstudents
like toread
P ( x ≥ 7) = 1 – P ( x < 7)
= 1 − 𝑃 (𝑥 = 6)
=1 - 9𝑐6 0.660.43
=1 – (84 )(0.046656) (0.064)
= 1- 0.25082
= 0.2318
P( x=7) + P ( x=8) + P (x=9)
9𝑐7 0.670.42 + 9𝑐8 0.680.41 +
0.01008
0.161243 + 9(0.0167616)(0.4)+
0.01008
0.1612+0.06034+ 0.01008
=0.2318

3. Cikgu Hanini Suhaila Hamsan Binomial Distribution puterizamrud@gmail.com
Determine mean,variance ,andstandard
deviationof abinomial distribution
µ =np α2=npq α =√ 𝑛𝑝𝑞 p+q=1
1 Givenn=200 p=
1
6
Findthe value q, µ, α2,α
p+q=1
q=1-p=
5
6
, µ =np
=200(
1
6
)
α2 = npq α =√ 𝑛𝑝𝑞
=200(
1
6
)(
5
6
) =√200(
1
6
)(
5
6
)
2 The fair dicesisrolled15 timescontinuously.
The probabilityof obtainingthe number2is
1
6
Find
a) The mean numberof timesthat
number2 appears
Mean =µ =np
= 15 (
1
6
)
b) The variance and the standard
deviationof number2isobtained
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝛼2 = npq =15(
1
6
)
5
6
) = 2.083
Standard Deviation=√2.083= =1.443
Johnyattempted60 questionswith4options
to choose from. There isonlyone correct
answerforeach questions.Johnyguessed all
the answers
Find
a) The mean numberquestionsthathe
will getright
Mean= np = 60 (
1
4
)= 15
b) Variance andthe standard deviation
of the numberof correct answer
Variance = npq= 15(
3
4
)= 11.25
Standarddeviation=√11.25 = 3.354
4
Giventhata classconsistsof 30 students,70
% of thempass ina mathematictest
Find
a) The mean numberof studentspass
the test
b) The variance and the standard
deviationof studentswhopassthe
test
a) Mean= µ =np
= 30 (0.7)
= 21
b) Variance = α2 = npq
= 30 (0.7)(0.3)
= 6.3
StandardDeviation = α =√ 𝑛𝑝𝑞
= √6.3
= 2.510