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AP Statistics
Chapter 8
Section 2
If you want to know the number of
successes in a fixed number of
trials, then we have a binomial
setting.
If you want to know how many
trials it will take to observe a
particular event, then we have a
geometric setting.
Vocab
• Geometric Setting
• Geometric distribution
• Mean or expected value of a geometric
random variable
Geometric Setting
1. “Success” or “Failure”
2. The probability of success is the same
for each observation
3. All observations are independent
4. The variable of interest is the number
of trials needed to obtain the first
success
An experiment consists of rolling a single die.
The random variable is defined as X = the
number of trials until a 3 occurs.
Geometric?
1. “3” or “not a three”
2. 1/6 probability of rolling a 3 each time
3. Independent events
4. Variable of interest – number of rolls until a 3
occurs
Yes, Geometric distribution
Rule for calculating Geometric
Probabilities
If X has a geometric distribution with p probability
of success and (1-p) of failure on each
observation, the possible values of X are 1, 2, 3,
…. If n is any one of these values, then the
probability that the first success occurs on the
nth trial is
P( X  n)  (1  p)
n 1
p
X
1
2
3
4
5
6
7
…
P(X)
p
(1-p)p
(1-p)2 p
(1-p)3 p
(1-p)4 p
(1-p)5 p
(1-p)6 p
…
p is the probability that the success will occur
on the 1st observation.
p(1-p) is the probability that the success will
occur on the 2nd observation.
…
Create the geometric distribution for the number
of rolls of a die until a 3 occurs.
p
1
6
1 1
p (1  p )  1    .1389
6 6
2
1 1
p (1  p ) 2  1    .1157
6 6
3
1 1
p1  p   1    .0965
6 6
...
3
X
P(X)
1
.1667
2
.1389
3
.1157
4
5
Calculator
• 2nd vars
• D geometric pdf, enter, (.16667, L1) sto L2
• L1 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …)
NO ZERO – it is not possible to find the
probability of the first success occurring on the 0
observation.
Look at the geometric pdf – this means that there
is approx. a 9.6% chance that the first time a 3 is
rolled will be the 4th roll of the die.
Calculator
What is the probability that it would take at
most 6 rolls of the die to produce a 3.
2nd vars
E Geometric cdf, enter, (.1667,L1) sto in L3.
There is approx a 67% chance that it would
take at most 6 rolls of the die to roll a 3.
Mean
1

p
The “expected value” or mean of the number
of rolls of a die to roll a 3 is
1
  6
1
6
The probability that it takes more than n trials to
see the first success is:
P X  n   1  p 
n
 1
P  X  4   1  
 6
4
4
5
    .4823
6
The probability that it will take more than 4
rolls of a die to roll a 3 is approx. 48%.