Transcript 18 The Geometric Distribution

“Teach A Level Maths”
Statistics 1
The Geometric
Distribution
© Christine Crisp
The Geometric Distribution
Statistics 1
OCR
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The Geometric Distribution
When playing some board games we have to roll a six
with a die before we can begin.
1
The probability of rolling a six with the 1 try is
.
6
st
The probability of not getting a six directly but
getting one with the 2nd try is
5 1

6 6
and failing twice but succeeding the 3rd time is
2
and so on.
5 5 1  5
1
    
6 6 6  6
6
The Geometric Distribution
In general, if X is the random variable “ the number
of trials to success”, and if the probability of
success is p and the probability of failure is q, where
p + q = 1, then
P ( X  1)  p
P ( X  2)  qp
2
q
P ( X  3)  p
and so on.
This distribution is called the geometric distribution.
The general term is given by P ( X  x )  q x 1 p
x 1
p
Substituting for q, we get P ( X  x )  (1  p)
In your formulae book this is written as p(1  p) x 1
The next page proves that X is a random variable
but if you haven’t studied the Geometric Series in
Pure Maths, skip over it.
SKIP
The Geometric Distribution
The sequence of probabilities for X = 1, 2, 3, . . . form
an infinite geometric sequence:
p , qp , q 2 p , . . .
The sum of an infinite number of terms is
a
S 
1 r
where a is the 1st term and r is the common
difference.
So, since a  p
and r  q
p
 S 
1 q
Since q = 1 p, we have 1 – q = p, so, S   1
The sum of the probabilities equals 1, the condition
for a random variable.
The Geometric Distribution
The Mean and Variance of the Geometric Distribution
The mean and variance of the Geometric Distribution
are given by
mean,
1

p
variance,  
2
1 p
p2
A random variable, X, with a geometric distribution
is written as
X ~ Geo( p)
The Geometric Distribution
Finding P ( X  x )
There are 2 ways of reasoning to get to the result.
1
e.g. Suppose we have X ~ Geo   and we want
 6
P ( X  5)
Method 1: P ( X  5)  1  P ( X  4)
 1  ( p  qp  q 2 p  q 3 p)
1
5
p
 q
so, using a calculator, we get
6
6
P ( X  5)  1  0  518
 0  482 ( 3 s. f . )
Method 2: The probability of success on the 5th trial,
or the 6th, or the 7th etc. means we have failure on the
1st 4, so, P ( X  5)  q 4. We get the same answer.
The Geometric Distribution
SUMMARY
 If X is a random variable “ the number trials to
success” then P ( X  1)  p
P ( X  2)  qp
2
q
P ( X  3)  p
and so on.
where, p is the probability of success and
q = 1 – p is the probability of failure.
 We write X ~ Geo ( p)
1
 The mean of X is given by  
p
1 p
2
 The variance of X is given by  
p2
( These results are in the formulae booklet )
 Also, P ( X  x )  q x 1
The Geometric Distribution
Exercise
1. When a darts player aims at a “double 20” on the
board the probability of hitting it is 0·2.
(a) What is the probability of hitting the double 20
(i) for the 1st time with the 3rd dart,
(ii) not hitting it in 10 tries?
(b) Find the expected number of throws before the
double 20 is hit.
Solution:
Let X be the random variable “ number of times
until the double 20 is hit”
Then, X ~ Geo (0  2)
(a)(i) P ( X  3) q 2 p  (0  8) 2 (0  2)  0  128
The Geometric Distribution
(a) What is the probability of hitting the double 20
(i) for the 1st time with the 3rd dart,
(ii) not hitting it in 10 tries?
(b) Find the expected number of throws before the
double 20 is hit.
X ~ Geo (0  2)
(a)(ii) We need P ( X  11)  q 10
 0  8 10
 0  107
1
(b) E ( X )   
p
5
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
The Geometric Distribution
In general, if X is the random variable “ the number
of trials to success”, and if the probability of
success is p and the probability of failure is q, where
p + q = 1, then
P ( X  1)  p
P ( X  2)  qp
2
q
P ( X  3)  p
and so on.
This distribution is called the geometric distribution.
The general term is given by P ( X  x )  q x 1 p
x 1
(
1

p
)
p
Substituting for q, we get P ( X  x ) 
In your formula book this is written as p(1  p) x 1
The Geometric Distribution
Finding P ( X  x )
There are 2 ways of reasoning to get to the result.
1
e.g. Suppose we have X ~ Geo   and we want
 6
P ( X  5)
Method 1: P ( X  5)  1  P ( X  4)
 1  ( p  qp  q 2 p  q 3 p)
1
5
p
 q
so, using a calculator, we get
6
6
P ( X  5)  1  0  518
 0  482 ( 3 s. f . )
Method 2: The probability of success on the 5th trial,
or the 6th, or the 7th etc. means we have failure on the
1st 4, so, P ( X  5)  q 4. We get the same answer.
The Geometric Distribution
SUMMARY
 If X is a random variable “ the number trials to
success” then P ( X  1)  p
P ( X  2)  qp
2
q
P ( X  3)  p
and so on.
where, p is the probability of success and
q = 1 – p is the probability of failure.
 We write X ~ Geo ( p)
1
 The mean of X is given by  
p
1 p
2
 The variance of X is given by  
p2
( These results are in the formula booklet )
 Also, P ( X  x )  q x 1