1. The document discusses several discrete probability distributions including the discrete uniform, hypergeometric, binomial, and Poisson distributions. 2. The hypergeometric distribution describes the probability of successes in a sample without replacement from a finite population with a known number of successes. 3. The binomial distribution gives the probability of a given number of successes in a sample of size n drawn with replacement from a population with probability p of success on each trial.

1. Probability and Statistics
Chapter 5
Some Discrete Probability
Distributions
Dr. Yehya Mesalam 1

2. Discrete uniform
2
k
i
k
i
X
P
k
U
X
,...,
2
,
1
1
)
(
)
,...,
2
,
1
(




1 2 3 k-1 k
1 2 3 k
Discrete uniform distribution
)
(x
f )
(x
F
0
k
1 k
2
1

k
k
k
1
Dr. Yehya Mesalam 2

3. Hypergeometric Distribution
The experiment consists of randomly drawing n elements
without replacement from a set of N elements, Np of
which are S’s (for success) and (N – Np) of which are F’s
(for failure).
Dr. Yehya Mesalam 3












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













n
N
X
n
Np
N
X
Np
n
Np
N
X
P )
,
,
(
N = Lot Size or Total number of elements
Np = Number of success in the N elements
n = Sample size
x = Number of success drawn in the n elements

4. Hypergeometric Distribution
N
Np
n
μ 
1
*
*




N
n
N
N
N Np
N
Np
n
2




























n
N
X
n
Np
N
X
Np
n
Np
N
X
P )
,
,
(
N = Lot Size or Total number of elements
Np = Number of success in the N elements
n = Sample size
x = Number of success drawn in the n elements
Dr. Yehya Mesalam 4

5. Example
• A manufacturing company uses an acceptance scheme
on production items before they are shipped. Boxes of
25 are readied for shipment and samples of 3 are tested
for defectives. If any defectives are found, the entire
box is sent back for 100% screening. If no defectives
are found, the box is shipped.
• a-What is the probability that a box containing 4
defectives will be shipped?
• b-What is the probability that a box containing only one
defective will be sent back for screening?
Dr. Yehya Mesalam 5

6. Solution
D
G
4
21
G
3
2
1
0
sent back for
100% screening
{ }box is shipped.
D
0
1
2
3
• n=3
• NP=4
• N=25
• X represent the defectives
0.578
3
25
3
21
0
4
0)
P(X 






















































n
N
X
n
Np
N
X
Np
X
P )
0
(
6
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7. Solution
D
G
4
21
G
3
3
1
0
sent back for
100% screening
{ }box is shipped.
D
0
1
2
3
• n=3
• NP=21
• N=25
• X represent the good
0.578
3
25
0
4
3
21
3)
P(X 






















































n
N
X
n
Np
N
X
Np
X
P )
3
(
7
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8. • n=3
• NP=1
• N=25
• X represent the defectives
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

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
















n
N
X
n
Np
N
X
Np
X
P )
0
( 
12
.
0
3
25
3
24
0
1
)
0
(
1 



























 X
P
Solution
8
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9. Binomial Distribution
!
( ) (1 )
! ( )!
x n x x n x
n n
p x p q p p
x x n x
 
 
  
 

 
p(x) = Probability of getting x ‘Successes’
n = Sample size
p = Probability of a ‘Success’ on a single trial
q = 1 – p ‫الحدث‬ ‫حدوث‬ ‫عدم‬ ‫او‬ ‫الفشل‬ ‫احتمال‬
) )
x = Number of ‘Successes’ in n trials
(x = 0, 1, 2, ..., n)
  E(x)  np   npq
Mean Standard Deviation
‫الحدث‬ ‫حدوث‬ ‫او‬ ‫النجاح‬ ‫احتمال‬
9
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10. Binomial Distribution
3 5 3
!
( ) (1 )
!( )!
5!
(3) .5 (1 .5)
3!(5 3)!
.3125
x n x
n
p x p p
x n x
p


 

 


Experiment: Toss 1 coin 5 times in a row. Note
number of tails. What’s the probability of 3 tails?
n = 5
p = 0.5 Probability of getting tail
q = 0.5
x = 3 Getting three tails
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11. Binomial Probability Table
n = 5 p
k .01 … 0.50 … .99
0 .951 … .031 … .000
1 .999 … .188 … .000
2 1.000 … .500 … .000
3 1.000 … .812 … .001
4 1.000 … .969 … .049
Cumulative Probabilities
p(x ≤ 3) – p(x ≤ 2) = .812 – .500 = .312
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12. n = p = x =
success =
Example
A marksman hits a target 80% of the
time. He fires five shots at the target. What is the
probability that exactly 3 shots hit the target?
3
3
3
)
3
( 

 n
n
q
p
C
x
P
5 .8
hit # of hits
3
5
3
)
2
(.
)
8
(.
!
2
!
3
!
5 

2048
.
)
2
(.
)
8
(.
10 2
3


12
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13. Example
What is the probability that more than 3 shots
hit the target?
5
5
5
5
5
4
5
4
5
4
)
3
( 



 q
p
C
q
p
C
x
P
0
5
1
4
)
2
(.
)
8
(.
!
0
!
5
!
5
)
2
(.
)
8
(.
!
1
!
4
!
5


7373
.
)
8
(.
)
2
(.
)
8
(.
5 5
4



13
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14. Example
The probability that a patient recovers from a rare blood
disease is 0.4. If 15 people are known to have contracted
this disease, what is the probability that
1-At least 10 survive?
X= number of people that survive
P(X ≥ 10) = 1 - P(X < 10) = 1 – P(X ≤ 9) = 1 – F(9) = 1 - 0.9662 = 0.0338
2-From 3 to 8 survive?
P(3 ≤ X≤8) = F(8) - F(2) = 0.9050 -0.0271 = 0.8779
3-Exactly 5 survive?
P(X=5)= F(5)-F(4)=0.4032-0.2173=0.1862
or
P(X =5) = f(5) = (0.4)5(0.6)10 = 0.1862
15
5
14
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15. 15
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16. Poisson Distribution
1. Number of events that occur in an interval
• events per unit
— Time, Length, Area, Space
2. Examples
• Number of customers arriving in 20 minutes
• Number of strikes per year in the U.S.
• Number of defects per lot (group) of DVD’s
16
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17. Poisson Distribution
p (x) = Probability of x given 
 = Mean (expected) number of events in unit
e = 2.71828 . . . (base of natural logarithm)
x = Number of events per unit
mean   
variance   
(x = 0, 1, 2, 3, . . .)
!
)
(
x
e
x
p
x




17
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18. Example
Customers arrive at a rate of
72 per hour. What is the
probability of 4 customers
arriving in 3 minutes?
© 1995 Corel Corp.
18
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19. Solution
72 Per Hr. = 1.2 Per Min.
 = 3.6 Per 3 Min.
 
-
4 -3.6
( )
!
3.6
(4) .1912
4!
x
e
p x
x
e
p



 
19
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20. Poisson Probability Table
x
 0 … 3 4 … 9
.02 .980 …
: : : : : : :
3.4 .033 … .558 .744 … .997
3.6 .027 … .515 .706 … .996
3.8 .022 … .473 .668 … .994
: : : : : : :
Cumulative Probabilities
p(x ≤ 4) – p(x ≤ 3) = .706 – .515 = .191
20
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21. Example
Suppose that the average number of telephone calls arriving
at the switchboard of a small corporation is 30 calls per hour.
1. What is the probability that 3 calls will arrive in a 1-minute period
30 calls ------ 60 min
? Calls -------- 1min
? = 30 = 0.5 f (3) = e -0.5 0.5 3
60 3!
2.What is the probability that no calls will arrive in a 3-minute period
0.5 calls ---- 1 min
? Calls ----- 3 min
?= 0.5 * 3 = 1.5 f(0)= e -1. 5 1.5 0 = 0.223
0!
3. What is the probability that more than five calls will arrive
in a 5 minutes interval ( Answer 0.042)
21
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22. Approximation
Binomial
Hypergeometric
Poisson Normal
N >> 50
P =
Np
N
__
q=1-p
=np
 =np
 =npq 22
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23. Example
Among the 120 applicants for a job only 80 are actually qualified. If
5 of these applicants are randomly selected for an interview, find the
probability that only 2 of the 5 will be qualified for the job by using
1. The hyper geometric distribution
80 40
2 3
120
5
X=2, n=5, N=120 ,Np=80
f(2)= = 0.164
2.The binomial distribution as an approximation.
2 3
5 2 1
2 3 3
f(2) = — — = 0.165
As can be seen these results are close .
P=80/120 =2/3
23
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24. Example
If 0.8% of the fuses delivered to an arsenal are defective, what is the
probability that four fuses will defective in random sample of 400?
The binomial distribution.
4 396
400
4
p= 0.008
q=0.992
n=400
0.008 0.992
f(4) = = 0.1788
The Poisson distribution as an approximation.
 = n*p = 400*0.008 = 3.2
178
.
0
!
4
2
.
3
)
4
(
4
2
.
3



e
p
24
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25. Example
Only one personal computer per thousand is
found to be defective after assembly in a
manufacturing plant, and the defective PCs are
distributed randomly throughout the production
run.
(a) What is the probability that a shipment of 500
PCs includes no defective computer?
(b) What is the probability that a shipment of 100
PCs includes at least one defective computer?
25
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26. b- P=0.001 q=0.999 =100(0.001)=0.1
Solution
09516
.
0
!
0
)
1
.
0
(
1
)
1
.
0
1
(
0
1
.
0






e
X
P 
a- By the Poisson approximation of binomial probabilities,
P=0.001 q=0.999 =500(0.001)=0.5
6065
.
0
!
0
)
5
.
0
0
(
5
.
0





e
X
P 
26
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27. 27
27
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