Poisson Probability Distribution
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Transcript Poisson Probability Distribution
Binomial Probability Formula
n
P ( x) C x p q
x
n x
n!
( n x )! x!
x
p q
n x
Binomial Probability Distribution
By listing the possible values of x with
the corresponding probability of each,
we can construct a Binomial Probability
Distribution.
Constructing a Binomial Distribution
In a survey, a company asked their workers and
retirees to name their expected sources of retirement
income. Seven workers who participated in the survey
were asked whether they expect to rely on Pension for
retirement income. 36% of the workers responded
that they rely on Pension only. Create a binomial
probability distribution.
Constructing a Binomial Distribution
x
P(x)
P ( 0 ) 7 C 0 ( 0 . 36 ) ( 0 . 64 ) 0 . 044
0
0.044
P (1) 7 C 1 ( 0 . 36 ) ( 0 . 64 ) 0 . 173
1
0.173
P ( 2 ) 7 C 2 ( 0 . 36 ) ( 0 . 64 ) 0 . 292
2
0.292
P ( 3 ) 7 C 3 ( 0 . 36 ) ( 0 . 64 ) 0 . 274
3
0.274
4
0.154
5
0.052
P ( 5 ) 7 C 5 ( 0 . 36 ) ( 0 . 64 ) 0 . 052
6
0.010
P ( 6 ) 7 C 6 ( 0 . 36 ) ( 0 . 64 ) 0 . 010
7
0.001
0
1
2
3
7
6
5
4
P ( 4 ) 7 C 4 ( 0 . 36 ) ( 0 . 64 ) 0 . 154
4
5
6
3
2
1
P ( 7 ) 7 C 7 ( 0 . 36 ) ( 0 . 64 ) 0 . 001
7
0
P(x) = 1
Notice all the probabilities are between 0 and
1 and that the sum of the probabilities is 1.
Population Parameters of a
Binomial Distribution
Mean: = np
Variance: 2 = npq
Standard Deviation: = √npq
Example
In Murree, 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.
Mean:
= np = 30(0.57) = 17.1
Variance: 2 = npq = 30(0.57)(0.43) = 7.353
Standard Deviation: = √npq = √7.353 ≈2.71
Problem 1
Four fair coins are tossed simultaneously. Find
the probability function of the random variable
X = Number of Heads and compute the probabilities
of obtaining:
No Heads
Precisely 1 Head
At least 1 Head
Not more than 3 Heads
Problem 2
If the Probability of hitting a target in
a single shot is 10% and 10 shots are fired
independently. What is the probability
that the target will be hit at least once?
Poisson Process
The Poisson Process is a counting that
counts the number of occurrences of
some specific event through time.
Number of
customers arriving to a
counter
Number of calls received at a telephone
exchange
Number of packets entering a queue
Poisson Probability Distribution
The Poisson probability distribution
provides a good model for the probability
distribution of the number of ‘rare events’
that occur randomly in time, distance, or
space.
Assumptions
Poisson Probability Distribution
The probability of an occurrence of an event
is constant for all subintervals and
independent events
There is no known limit on the number on
successes during the interval
As the unit gets smaller, the probability that
two or more events will occur approaches
zero.
µ=1
µ=4
µ = 10
Poisson Probability Distribution
e
x
f ( x)
,
for x 0, 1,2,...
x!
• f(x) = The probability of x successes over a
given period of time or space, given µ
• µ
= The expected number of successes per
time or space unit; µ > 0
• e
= 2.71828 (the base for natural logarithms)
Problem 5
Let X be the number of cars per minute
passing a certain point of some road between 8
A.M and 10 A.M on a Sunday. Assume that X
has a Poisson distribution with mean 5. Find
the probability of observing 3 or fewer cars
during any given minute.
Problem 7
In 1910, E. Rutherford and H. Geiger showed
experimentally that number of alpha particles
emitted per second in a radioactive process is
random
variable
X
having
a
Poisson
distribution. If X has mean 0.5. What is the
probability of observing 2 or more particles
during any given second?
Problem 9
Suppose that in the production of 50 Ω resistors,
non-defective items are those that have a resistance
between 45 Ω and 55 Ω and the probability of being
defective is 0.2%. The resistors are sold in a lot of 100,
with the guarantee that all resistors are non-defective.
What is the probability that a given lot will violate this
guarantee?
Problem 11
Let P =
1%
be the probability that a
certain type of light bulb will fail in 24
hours test. Find the probability that a sign
consisting of 100 such bulbs will burn 24
hours with no bulb failures.
Multinomial Distribution
If a given trial can result in K outcomes E1,E2, …, Ek
with probabilities p1,p2, …,pk, then the Probability
Distribution of the random variables X1,X2, …, Xk,
representing the number of occurrences for E1,E2, …, Ek
in n independent trials is
n
x1 x 2
p1 p 2 ... p kxk
f ( x 1 , x 2 , , x k ; p 1 , p 2 , , p k , n)
x1 , x 2 ,..., x k
k
x
i
n
i 1
k
i 1
pi 1
Example
An airport has three runways. The probabilities that
the individual runways are accessed by a randomly
arriving commercial jets are as following:
Runway 1: p1 = 2/9
Runway 2: p1 = 1/6
Runway 3: p1 = 11/18
What is the probability that 6 randomly arriving
airplanes are distributed in the following fashion?
Runway 1: 2 airplanes
Runway 2: 1 airplanes
Runway 3: 3 airplanes
Sampling With Replacement
x
n M
M
f ( x )
1
N
x N
N all items
M defective
p
M
(Pr obability )
N
n trials
n x
Hypergeometric Probability Distribution
In cases where the sample size is relatively
large compared to the population, a discrete
distribution called hypergeometric may be
useful.
Sampling Without Replacement
Hypergeometric Distribution
M
f (x)
x
N M
*
n x
N
/
n
N
n
= Different ways of picking n things from N
M
x
= Different ways of picking x defective from M
N M
n x
= Different ways of picking n-x
nondefective from N-M
Hypergeometric Distribution
Mean and Variance
Mean n
M
N
Variance
2
nM ( N M )( N n )
N ( N 1)
2
Problem 13
Suppose that a test for extra sensory
perception consists of naming (in any
order) 3 cards randomly drawn from a
deck of 13 cards. Find the probability
that by chance alone, the person will
correctly name (a) no cards, (b) 1 Card,
(c) 2 Cards, and (d) 3 cards.
Quiz # 2
32 Cptr (B) – 5 NOV 2012
If the Probability of hitting a target in
a single shot is 5% and 20 shots are fired
independently. What is the probability
that the target will be hit at least once?