6.3 The Geometric Distribution
Objectives
 Recognize a geometric distribution.
 Evaluate geometric probabilities using the formula and calculator.
 Calculate the mean (expected value) and variance of a geometric distribution.
The geometric distribution allows you to model and find probabilities for the number of trials needed
until the first success occurs.
Examples of geometric distribution problems:



Find the chance you will need to buy more than ten instant-win lottery tickets to win a prize.
Find the chance of ten days passing before the first accident occurs at a certain intersection.
Find the chance a typist can go no more than ten pages without making an error.
The geometric random variable X is the number of trials necessary to obtain the first success. It can take
on the values 1, 2, 3, …
(Note: Unlike the binomial random variable it cannot be zero, as the earliest the first success can come is
in the first trial.)
Geometric Conditions
A probability problem must meet the following four conditions for the random variable, X, within the
problem to have a geometric distribution:
1. Each trial can result in one of two possible outcomes, “success” or “failure”.
2. The trials are independent.
3. The probability of success is the same for all trials.
Finding Probabilities
If X is the number of trials up to and including the first success with the probability p of success on each
trial and the possibility of failure 1 – p, the possible values of X are 1, 2, 3, … (which means the number
of failures prior to the first success is X – 1). If x is any one of these values, the probability that the first
success occurs on the xth trial is
P( X  x)  (1  p) n1 p
This formula is quite simple and doesn’t really require the use of the geometpdf(p,x)formula in the
DISTR menu on your calculator.
1
6.3 The Geometric Distribution
Note on solving “At Least” problems:
These can be computed in two different ways:
P( X  k)  1  P( X  k)  1   P( X  1)  P( X  2)  ...  P( X  k  1) 
or
P( X  k)  (1  p) k 1 p  (1  p) k p  (1  p) k 1 p  ....
 p (1  p) k 1  (1  p) k  (1  p) k 1...
 (1  p) k 1 
a
 p
(using the G.S. formula for a sum to infinity
)

1 r
 1  (1  p) 
 (1  p) k 1 
 p

p



 1 p

k 1
Examples:
1. Suppose that 40% of the students who drive to your university campus carry jumper cables. Your car
has a dead battery and you don’t have jumper cables, so you decide to stop students and ask them
whether they have a pair of jumper cables. Find the probability that three or fewer students must be
stopped.
2. A die is rolled until a 3 is observed. The probability that it takes more than 6 rolls to observe a 3 is:
3. Before starting to play the game “Snakes and Ladders” each player throws a die until a six is
obtained. The number of throws before a player starts is the random variable Y, where Y takes the
values 1, 2, 3, … Two people play Snakes and Ladders. Calculate the probability that they will each
need at least five throws before starting.
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6.3 The Geometric Distribution
Shape, Center and Spread
All geometric distributions are strongly skewed to the right because as n increases we are multiplying by
a number less than one, 1 – p, therefore the probabilities become less likely.
1
p
(1 p) q
Var( X )   2X 
 2
p2
p
E( X )   X 
However, the different geometric distributions differ in their mean and spread. As p increases, the mean
and the spread both decrease.
Example: Glenn likes the game at the state fair where you toss a coin into a saucer. You win if the coin
comes to rest in the saucer without sliding off. Glenn has played this game many times and has
determined that on average he wins 1 out of every 12 times he plays. He believes that his chances of
winning are the same for each toss. He has no reason to think that his tosses are not independent. Let X be
the number of tosses until a win.
1

p
1 p
Var( X )  2 
p
E( X ) 
SD( X )  Var( X ) 
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6.3 The Geometric Distribution