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ENGG2450 Probability and Statistics for Engineers
1
Introduction
3
Probabilityy
4
Probability distributions
5
Probability densities
2
Organization and description of data
6
Sampling distributions
7
Inferences concerning a mean
8
C
Comparing
two treatments
9
Inferences concerning variances
A
Random Processes
4 Probability distributions
4.1 Random variables
4.2 The binomial distribution
4.3 The hypergeometric distribution
1 Introduction
3 Probability
4 Probability distributions
5 Probability densities
2 Organization & description
6 Sampling distributions
7 Inferences .. mean
8 Comparing 2 treatments
9 Inferences .. variances
A Random processes
4 4 The mean and the variance of a
4.4
probability distribution
4.5 Chebyshev’s Theorem
4 6 The Poisson approximation to the
4.6
binomial distribution
4.8 The geometric and negative binomial distribution
4.1 Random variables
In a
an e
experiment,
pe e t,
there is an
outcome.
An event A is said to have
occurred in an experiment if the
outcome is an element of A.
e.g.1 Throwing a dice
outcome
xi  S
(3)
Sample
p space
p
is the set
corresponds to all possible
outcomes of an experiment.
sample space
S={1,2,3,4,5,6}
Suppose the outcome of an
experiment is a number. We
can define a quantity called
In this experiment
p
of throwing
ga
dice, the outcome is a number
random variable X to
sample of the random variable X.
represent the randomness
nature of the number
number.
The i in xi represents the i trial
represented here by xi which is a
of the experiment.
In an experiment,
there is an outcome.
The outcome of an experiment
is a number and is a sample of
the random variable.
e.g.1 Throwing a dice
outcome
Sample space is the set
corresponds to all possible
outcomes of an experiment.
p
xi  sample space S ={1,2,3,4,5,6}
The result of throwing a dice is random and discrete. It can be
represented by a discrete random variable, say X.
We obtain a sample of the random variable, represented as xi , by throwing
a dice, where i indicates that the sample is obtained in the i th experiment.
i is discrete.
e.g.2 Measuring noise
level of a phone line
x(t)
outcome x(t)  sample space (-10v, 10v)
The noise level of a phone line is random and continuous. It can
be represented
p
by
y a continuous random variable,, sayy X.
We obtain a sample of the random variable, represented as x(t) , by measuring
the noise level of a phone line at certain time t.
t
is continuous.
4.1 Random variables
(5)
A random variable is defined by the following:
1. An experiment that can be repeated.
The set that contains all possible outcomes of the experiment is called the sample space. A
subset of the sample
p space
p
is called an event.
The outcome of an experiment is a number which is a sample of the random variable.
2. A function that associates a probability to each event.
Experiment
e.g.1
obtain a sample of the
random variable
outcome  sample
space
Throw a dice at trial i and
xi {1,2,3,4,5,6}
record the outcome
e.g.2
xi.
Measure the noise level of a
phone line at time t and record
the outcome x(t).
voltage
x(t)  (-10v, 10v)
Note
ote that
t at in e.g.
e.g.2, the
t e outcome
outco e x(t) iss continuous,
co t uous, so tthere
eea
are
e infinite
te number
u be o
of
possible outcome. What will be the probability of each of this outcome?
The probability of a possible outcome is 0.
P(1 ≤ X ≤ 3) may not be 0.
A random variable is any function that assigns a numerical value to
each possible outcome.
The probability distribution of a discrete random variable X is a list of
the possible values of X together with their probabilities f(x) = P[ X=x].
The probability distribution always satisfies the conditions
f(x)  0 and

f(x) =1.
all x
e.g.
f(1) = P[ X=1]= 1/6 = f(2) = f(3)= f(4)= f(5)= f(6)
The cumulative distribution function or distribution function of a
discrete random variable X is F(x)
F( ) = P[ X ≤ x].
]
e.g.
F(1)
( ) = P[[ X1]=
] 1/6
F(3) = P[ X3]= 3/6
F(2)
( ) = P[[ X2]=
] 2/6
F(6) = P[ X6]=
1
4.2 The Binomial Distribution
Bernoulli trials
(7)
1. There are only two possible outcomes, called success & failure.
2. The probability of success, denoted by p, is the same for each trial.
note:
The probability
of failure is 1-p.
3. The outcomes from different trials are independent.
p 4,, we further assume
In the pproblems to be studied in chapter
4. A fixed number n of Bernoulli trials are conducted.
eg
e.g.
Consider
C
id th
the ttrial
i l off th
throwing
i a di
dice with
ith success defined
d fi d as h
having
i an
outcome 6.
The probability of success is p =1/6.
Suppose the dice is thrown 4 times. Can these 4 trials be
treated as Bernoulli trials? Yes.
Yes
What are the possible results of these trials?
They are
ssss
fsss
sffs
fffs
sssf
fssf
ssfs
fsfs
ssff
fsff
sfss
ffss
sfsf
ffsf
sfff
ffff
They are 16 possible results in these Bernoulli trials with n = 4 .
n= 4
ssss sssff ssfs
f ssff
ff sfss
f sfsf
f f sffs
ff sfff
fff
fsss fssf fsfs fsff ffss ffsf fffs ffff
x cn,x
X to
t representt the
th number
b
of success in this experiment. The sample space is {0,1,..., n}.
Let x be a sample of X.
X
W use discrete
We
di
t random
d
variable
i bl
The probability of a result with x successes
(which implies n – x failures) is px (1-p)n-x.
The probability of all results with x successes is
f ( x)  P[ X  x]  Cn , x  p x  q n  x 
n!
 p x  q n x
x ! n  x  !
The probability distribution of
X is Binomial distribution
b( x; n, p)  Cn, x  p x  (1  p)n x
x  0, 1, 2,..., n
0
1
1
4
2
6
3
4
4
1
n= 5
x cn,x
0
1
1
5
2
10
3
10
4
5
5
1
e.g. It has been claimed that in 60% of all solar-heat installations the
utility bill is reduced by at least one-third. Accordingly, what are the
probabilities that the utility bill will be reduced by at least one-third in
(a) four of five installations;
(b) at least four of five installations?
sln. Consider the trial of building a solar-heat installation.
Success is defined as having the bill reduced by at least 1/3.
Probability of success p(s) = p = 0.6
The probability of x
successes in n
Bernoulli trials
b(x;n, p)  Cn,x  px  qnx
The outcomes from different trials are independent
and so they are Bernoulli trials.
x  0,1, 2,...,n
(a) four of five installations are successful,
successful ii.e.
e n=5,
n=5 x=4
b( x; n, p )  C5, 4  p 4  (1  p )5 4
= C5,4  0.64  (1-0.6) = 0.259
(b) at least four of five installations are successful, i.e. n=5, x=5 & 4
b( x; n, p )  C5 , 5  p 5  (1  p )55
0 078
= C5,5  0.6
0 65  (1-0.6)
(1-0 6)0 = 0.078
The probability is b(4; 5, 0.6) + b(5; 5, 0.6) = 0.259 + 0.078 = 0.337
4.2 The Binomial Distribution
(10)
Binomial distribution
b( x; n, p)  Cn, x  p x  (1  p)n x
x  0, 1, 2,..., n
4.1 Random variables
4.2 binomial distribution
4.3 hypergeometric
distribution
4.4 mean & variance
4.5 Chebyshev’s Theorem
4.6 Poisson approximation
4.8 geometric & negative
binomial distribution
If n is large, the calculation of binomial probabilities become tedious.
A binomial distributions table gives the values of cumulative probabilities
x
B( x; n, p)   b( x; n, p)
x  0, 1, 2,..., n
k 0
for n =2 to 20 and
p = 0.01,
0 01 0.05,
0 05 010,
010 0.15,
0 15 ..., 0.90,
0 90 0.95.
0 95
Cumulative probabilities B(x;n,p) rather than probabilities b(x;n,p) are
tabled because the values of B(x;n,p)
B(x;n p) are the ones needed more often
in statistical applications.
P b biliti b(x;n,p)
Probabilities
b(
) can be
b d
determined
t
i d ffrom cumulative
l ti probabilities
b biliti
b(x;n,p) = B(x;n,p) - B(x-1;n,p) where B(-1;n,p)=0.
e.g. A manufacturer of external hard drives claims
that only
y 10% of his drives require
q
repairs
p
within
the warranty period of 12 months. If it turns out that
5 of 20 of his drives requires repairs within the first
year, does this tend to support or refute the claim?
Binomial distribution
b ( x; n, p )  Cn , x  p x  (1  p ) n  x
x  0,1,..., n
sln. Let p probability of success (repairs needed in 12 mths) = 0.1
The probability that 5 or more of 20 of the drives will require repairs is
b(5; 20,p) + b(6; 20,p) + ... +
b(20; 20,p)
= 1 - { b(0; 20, p) + b(1; 20, p) + ... + b(4; 20, p) }
= 1-
B(4 20,p)
B(4;
20 )
= 1 - 0.9568
= 0.0432
0 0432
Since this p
probability
y is very
y small, it would seem reasonable to reject
j
the hard drive manufacturer’s claim.
= B(x; n, p) = B(4; 20, 0.10)
Examples of Binomial Distributions
b(x;n,p)
Binomial distribution
b ( x; n, p )  Cn , x  p x  (1  p ) n  x
x  0,1,..., n
x
For any n, the binomial distribution with p=0.5 is a symmetrical
distribution at x= n/2.
b(x;n,p)
When p<0.5,
p
X is more
likely to be
small.
When p>0.5,
X is more
likely to be
large.
x
4.3 The Hypergeometric Distribution
(15)
Consider a lot consisting of N units,
units of which a are defective.
defective
e.g. N= 20, a=3
1 2 3 4 5 6 7 8 9 10 11121314151617181920
We are interested in the number of defectives in a sample of n units taken
? ? ? ? ? ? ? ? ? ?
from this lot. e.g. n = 10
p be drawn in such a way
y that at each successive drawing,
g,
Let the sample
whatever units left in the lot have the same chance of being selected.
Suppose we do sampling without replacement.
The probability the 1st unit drawn from the sampled n units is defective is a/N.
The probability the 2nd unit drawn is defective is (a-1)/(N-1) or (a)/(N-1),
depending on whether the 1st unit drawn is defective or not defective.
The trials are not independent and so not Bernoulli trials. The probabilities
of drawing a defective unit cannot be determined using Binomial distribution.
The trials are independent and so are Bernoulli trials if we do sampling
with replacement.
Consider a lot consisting of N units, of which a are defective.
e.g.
g N= 20,, a=3
1 2 3 4 5 6 7 8 9 10 11121314151617181920
We are interested in the number of defectives in a sample of n units from this lot.
e.g. n = 10, x = 2 ?1?2?3?4?5?6?7?8 ?18?19; ……
We do sampling without replacement.
Suppose there are x defective units in this sample of n units.
These x defective units can be chosen in Ca,x ways.
The n-x non-defective units can be chosen in CN-a,, n-x ways.
There are Ca,x x CN-a, n-x ways.
There are
CN,n ways to choose n units from the N units.
units
Hypergeometric distribution
 a  N  a 
 x  n  x 
 , x  0, 1, ..., n
h( x; n, a, N )    
N
n
 
The probability of a result with x
successes (defective) in n trials is
Hypergeometric distribution.
e.g. A company sells discount accessories for cell phones often ships many
defective units. Suppose it has 20 chargers on hand but 5 are defective.
If the company decides to randomly select 10 of these items, what is the
probability that 2 of the 10 is defective.
sln. N = 20, a = 5,
n = 10,
10 x = 2.
2
1 2 3 4 5 6 7 8 9 10 11121314151617181920
P( 12345678 1617 …..))
The probability that 2 of the 10 will be defective
 5  15 
 2 8 
 h( x; n, a, N )     
 20 
 10 
 
= (10 x 6,435 )/184,756 = 0.348
sampling without
replacement.
p
Hypergeometric
distribution  a   N  a 
 x  n  x 
,
h( x; n, a, N )    
N
n
x = 0, 1, ..., n.
 
e.g. A company sells discount accessories for cell phones often ships many
defective units. Suppose it has 100 chargers on hand but 25 are defective.
The company decides to randomly select 10 of these items.
What is the probability that 2 of the 10 is defective
using (a) hypergeometric distribution and
(b) binomial distribution as an approximation?
sln.
l The
Th probability
b bilit that
th t 2 off the
th 10 will
ill b
be d
defective
f ti
hypergeometric
distribution  a   N  a 
 x  n  x 
,
h( x; n, a, N )    
N
n
x = 0, 1, ..., n.
 
(a) hypergeometric distribution N = 100, a = 25, n = 10, x = 2.
C  C75 ,8
h( x; n, a, N )  25 , 2
= 0.292
C100 ,10
(b) binomial distribution as an approximation p = 25/100 = 0.25,
0 25 n = 10,
10 x = 2.
2
b( x; n, p )  C10 , 2  p 2  (1  p )10 2= 0.282
It can be shown that h(x;n,a,N)
h(
) approaches
b(x;n,p) when N .
binomial distribution
b( x; n, p)  Cn, x  p x  (1  p) n  x
A good rule of thumb: binomial distribution is a good approximation to
hypergeometric distribution when N  10 n .
Examples of Hypergeometric Distributions
h(x; n,
n a,
a N)
hypergeometric
distribution  a   N  a 
 x  n  x 
,
h( x; n, a, N )    
N
n
x = 0, 1, ..., n.
 
x
b(x; n, p)
x
4.4 The Mean and Variance of a Probability Distribution
(20)
Mean of discrete p
probability
y distribution f(
f(x))
   x  f (x)
all x
A Bernoulli trial is an experiment that has only 2
outcomes, success & failure. p(success)= p; p(failure)=1-p.
Outcomes from different trials are independent
independent.
Consider a coin is tossed 3 times. Find the mean number of heads ((success).
)
The number of heads in 3 tosses is a discrete random variable, say X.
There are 8 possible results in these 3 Bernoulli trials.
trials
sss ssf sfs sff fss fsf ffs fff
1. x is a sample of X.
2. The sample space of X is {{0,, 1,, 2,, 3 }
3. f(x) is probability distribution of X & is binomial.
4. Mean of f(x) is also called Mean of X or expected
value of X
n=3
x
cn,x
0
1
1
3
2
3
3
1
e.g. Find the mean number of heads (success) in 3 tosses.
sln. Let X be the number of heads which is a discrete
random variable of binominal distribution.
The probability of success p = 0.5
There are 8 possible results in these 3 Bernoulli trials.
sss ssf sfs sff fss fsf ffs fff
The mean number of heads  
 x  f (x)
all x
3
  x  b( x; n, p)
n=3
x
cn,x
0
1
1
3
2
3
3
1
binomial distribution
b( x; n, p )
 Cn , x  p x  (1  p ) n  x
x 0
= 0 x b(0;3,p) + 1 x b(1;3,p) + 2 x b(2;3,p) + 3 x b(3;3,p)
= 0 + 1 x 3 x 0.51 x ((1-0.5))2 + 2 x 3 x 0.52 x ((1-0.5)) + 3 x 1 x 0.53 x ((1-0.5))0
= 0 + 0.375 + 0.75 + 0.375 = 1.5
Alternatively,  
3

x 0
x  b( x; n, p)  n p = 3 x 0.5 = 1.5
Mean of binomial distribution f(x)
Mean of f(x)
   x  f (x)
  n p
all x
x  f ( x)
Pf.   
all x

  x  Cn, x  p x  (1  p)n x
binomial distribution
b ( x; n , p )  C n , x  p x  (1  p ) n  x

all x
n
n!
  x
 p x  (1  p)n x
x!(n  x)!
x 1
n!
 p x  (1  p)n x
) (n  x)!
)
x 1 ( x  1)!
n

n
 np 
x 1
m
 np 
y 0
(n  1)!
 p x 1  (1  p) n x
( x  1)!(n  x)!
Let y = x-1,, m = n-1
m
m!
y
m y
 p  (1  p)  n p  b( y; m, p)
y!(m  y)!
y 0
=np
Mean of hypergeometric distribution f(x)
a
  n
N
   x  f (x)
all x
Pf.    x  f (x)
all x
Mean of f(x)
hypergeometric
distribution
:
You will prove it in Assignment 2.
x = 0,
a N  a
  

x   n  x 

h ( x; n, a , N ) 
,
N 
 
1, ..., n.
 n 
e.g. A company sells
ll di
discountt accessories
i ffor cellll phones
h
often
ft ships
hi many
defective units. Suppose it has 20 chargers on hand but 5 are defective.
If the company decides to randomly select 10 of these items. What is the
mean of the probability distribution of the number of defective?
sln N = 20,
sln.
20 a = 5,
5
n = 10
  n
1 2 3 4 5 6 7 8 9 10 11121314151617181920
? ? ? ? ? ? ? ? ? ?
a
5
 10   2.5
N
20
Variance of probability distribution f(x)
 2   ( x   )2  f ( x)
 is called the standard deviation
which measures the expected
deviation of samples from the mean.
all x
Probab
bility
Binomial distributions
b( x; n, p )  Cn , x  p x  (1  p ) n  x
x = 0,
0 11, ..., n
 = 5/4
x
20/4
50/4
Variance of binomial distribution
 2  n  p  (1  p )
Variance of probability distribution f(x)
 2   ( x   )2  f ( x)
 is called the standard deviation
which measures the expected
deviation of samples from the mean.
all x
x
Variance of hypergeometric distribution
 2  n
a 
a  N  n 
 1  

N
N  N  1 
Hypergeometric
distribution
 a  N  a 
 x  n  x 
,
h( x; n, a, N )    
N
n
 
x = 0, 1, ..., n.
Mean of f(x)
k th moment about the origin
   x  f (x
( x)
   x  f (x)
k
'
k
all x
all x
 1'
 1st moment about the origin
Variance of f(
f(x))
k th moment about the mean
 k   ( x   )k  f ( x)
 2   ( x   )2  f ( x)
all x
all x
 2
 2 nd moment about the mean
C
Computing
ti fformula
l for
f the
th variance
i
 2   2'   2
4.5 Chebyshev’s Theorem
(27)
Theorem 1: If a probability distribution has mean  and standard
deviation , the probability of getting a value which deviates from
1
 by
b att least
l
t k
k is
i att mostt 1/k2 , i.e.
i
P | X   | k   2 .
k
ee.g.
g The number of customers
who visit a car dealer’s
showroom is a random variable
with
ith =18
18 and
d =2.5.
2 5 With
what probability can we assert
that there will be more than 8
but fewer than 28 customers?
sln.
| x   | k
P | X   | k  
f( )
f(x)

4
(   x )  k or ( x   )  k

{ 8 or less } or { 28 or more } customers
4
1
1

k 2 16
Probability of having more than 8
& fewer than 28 customers
 P | X   | k   1 
1
15
or
16 16
e.g.
w4
Show that for 40,000 flips of a
balanced coin, the probability is at
least 0.99
0 99 that the proportion of heads
will fall between 0.475 and 0.525.
Chebyshev’s Theorem:
1
P | X   | k    2 .
k
sln. Let discrete random variable X be the
number of heads in 40,000 flips.
These are Bernoulli trials and so
of binominal distribution.
f(x)
X is
k
= n p = 40,000 x ½ = 20,000
1 1
 100
2 2
  np (1  p )  40000  
binomial distribution
= n p ;
P  | X   | k   0.99
Byy Chebyshev’s
y
Theorem, 1 
1
 0.99
k2
k
so k = 10
It means X has at least 0.99 chance within (kk
or (20000-1000, 20000+1000)
or a proportion within (19,000/40,000, 21,000/40,000).
 = n p (1-p)
Slide 28
w4
wkcham, 21/01/2016
4.6 The Poisson approximation to the binomial distribution
(29)
Poisson distribution
f ( x; ) 
x e
x!
x  0, 1, 2,...,   0
binomial distribution
b( x; n, p )  Cn , x  p x  (1  p ) n  x
The Poisson distribution often serves as a distribution model for counts
which do not have a natural bound.
For example, an individual keeping track of
the amount of mail they receive each day
may notice that they receive an average
number of 4 letters per day.
If receiving any particular piece of mail
doesn't affect the arrival times of future
pieces of mail, i.e., then a reasonable
assumption is that the number of
pieces of mail received p
p
per day
y obeys
y
a Poisson distribution.
f(x,)
x
When n is large and p is small, binomial
probabilities are often approximated by means
of Poisson distribution with  = np.
Prove
lim
b( x; n, p)  Cn, x  p x  (1  p)n x  f ( x; )
n 
Poisson distribution
f ( x;  ) 
x!
, x  0, 1,,...,,   0
binomial distribution
b( x; n, p)  Cn, x  p x  (1  p)nx
x = 0, 1, ..., n
x
lim
lim
n!


Pf:
f
b( x; n, p) 
    (1  ) n x
n   x!(n  x)!  n 
n
n

x e 
lim n(n 1)(n  x  1) x
 n x



(
1

)
x
n 
x! n
n
1
2
x
11
)
(
1

)(
1

)

(
1

lim

n
n
n
 x  (1  ) n x

n
x!
n
lim x


 (1  ) n  x
n   x!
n
1
2
x 1
(1 )(1 )(1
) 1
n 
n
n
n
lim
When n is large and p is small, binomial probabilities are often
approximated by means of Poisson distribution with  = np.
Prove
lim lim
b( x; n, p)  Cn, x  p x  (1  p)n x  f ( x; )
n  p 0
Poisson distribution
f ( x;  ) 
x e 
x!
, x  0, 1,,...,,   0
binomial distribution
Pf: (continued)
lim x

b( x; n, p) 
 (1  ) n x
n
n   x!
n
lim

x e  
x!
, x  0, 1,...,   0
b( x; n, p)  Cn, x  p x  (1  p)nx
x = 0, 1, ..., n
lim


(1  )n (1  ) x  e
n 
n
n
which is Poisson distribution
A rule of thumb is to use this approximation if n ≥ 20, p ≤ 0.05.
If n ≥100, the approximation is excellent as np ≤ 10.
e.g. A heavy machinery manufacturer has
3840 large
g g
generators in the field that are
under warranty. If the probability is 1/1200
that any one will fail during the given year,
find the probabilities that 0,1,2,3,…
0123
of the
generators will fail during the given year.
sln:
Poisson distribution
f ( x;  ) 
x e 
x!
, x  0, 1,...,   0
binomial distribution
b( x; n, p)  Cn , x  p x  (1  p) n  x
x = 0, 1, ..., n
A n ≥100 and
As
d np ≤ 10, Poisson
P i
approximation
i
i iis valid.
lid
The probability is f ( x;  ) 
x e 
x!
, x  0, 1, 2,...
where  = np = 3840 · 1/1200
= 3.2
Mean of Poisson distribution
  
V i
Variance
off Poisson
P i
distribution
di t ib ti

2

mean of binomial distribution
 = n p
variance of binomial distribution
 = n p (1- p)
4.8 The geometric and negative binomial distribution
(33)
Geometric distribution
g(x; p) = p ( 1- p
)x-1
for x = 1, 2, 3, 4, …
e.g. Cars manufactured by a company need to be
checked if the nitrogen oxide emission meets
governmentt standard.
t d d S
Suppose we are
interested in the number of cars people have to
inspect until they find one whose emission does
not meet the standard (success).
4.1 Random variables
4.2 binomial distribution
4.3 hypergeometric
distribution
4.4 mean & variance
4.5 Chebyshev’s Theorem
4.6 Poisson approximation
4.8 geometric & negative
binomial distribution
If the
th first
fi t success is
i to
t come on the
th xth trial,
t i l th
then th
the probability
b bilit iis
p ( 1- p )x-1 where p is the probability of success.
Mean of geometric distribution
 
1
p
4.8 The geometric and negative binomial distribution
The number of Bernoulli trials tried to find
the first success is geometric distribution.
The number
Th
b off Bernoulli
B
lli ttrials
i l ttried
i d tto fifind
d
r successes is
Negative binomial distribution
 x  1 r
 p (1  p) x r
f ( x)  
1
 r 1
for x = r,, r+1,, …
Th rthh success occurs att the
The
th xthh Bernoulli
B
lli ttrials.
i l
There are r-1 successes in the previous x-1 trials.
The probability
f(x) = b(r-1; x-1, p) p
 C x 1,r 1  prr  (1  p ) x  r
(34)
Geometric distribution
g(x; p)= p (1-p)x-1
Probability of y successes
in n Bernoulli trials
b( y; n, p)  Cn, y  p y  (1  p) n y
y  0,1, 2,..., n
e.g. 3 Bernoulli trials
h
have
8 possible
ibl results.
lt
sss ssf sfs sff fss fsf ffs fff
Do’s and Don’ts
(35)
Do s
Do’s
Describe the chance behavior of a discrete random variable X by its
probability distribution function f(x) = P(X=x) for each value of X in
the sample space.
Summarize a probability distribution f(x) or the random variable X by
its mean  and variance .
Binomial distribution,
distribution hypergeometric distribution,
distribution Poisson distribution
have their own sets of underlying assumptions. Choose the right
distribution for your problem.
Don’ts
Never apply the binomial distribution to counts without first checking that
the conditions hold for Bernoulli trials.