1. The Geometric Probability Distribution
Skill 26

2. Objectives
Use the geometric distribution to compute the probability that the nth trial is the first success.
Apply the geometric distribution to real examples.

3. Geometric Distribution
The geometric probability distribution is the probability that our first success comes on the nth trial.
Question you will be asked to solve…
What is the probability that our first success comes on the nth trial?

4. n = # when 1st success happens
p = P(Success)

5. Example–Geometric Probability
An automobile assembly plant produces sheet-metal door panels. Each panel moves on an assembly line.
As the panel passes a robot, a mechanical arm will perform spot welding at different locations. Each location has a magnetic dot painted where the weld is to be made.
The robot is programmed to locate the magnetic dot and perform the weld. However, experience shows that on each trial the robot is only 85% successful at locating the dot.

6. Example–Geometric Probability
The robot will keep trying until it finds the dot (and does the weld) or the door panel passes out of the robot’s reach.
What is the probability that the robot’s first success will be on attempts n = 1, 2, or 3?
Solution:
Since the robot will keep trying until it is successful, the geometric distribution is appropriate.

7. Example–Solution
In this case, success S means that the robot finds the correct location.
The probability of success is p = P(S) = 0.85.
The probabilities are
n P (n) = p(1 – p)n – 1 = 0.85(0.15)n – 1
(0.15)0 = 0.85
(0.15)1 =
(0.15)2 

8. Example–First success
The assembly line moves so fast that the robot has a maximum of only three chances before the door panel is out of reach. What is the probability that the robot will be successful before the door panel is out of reach?
Solution:
n = 1 or 2 or 3 are mutually exclusive
P (n = 1 or 2 or 3) = P(1) + P(2) + P(3)  =
This means that the weld should be correctly located about 99.7% of the time.

9. Example–Geometric Probability
a. What is the probability that the robot will not be able to locate the correct spot within three tries?
P(robot will correctly locate the weld) =
P(not locate the weld after three unsuccessful tries)
1 – =
b. If 10,000 panels are made, what is the expected number of defectives?
(10,000)(0.0034) = 34 defectives.

10. Example–Geometric Probability
Suppose you are on the trap shooting team, you have an 80% chance of hitting the target with each shot. Find the probability you will hit the target for the first time on the second attempt.
Solution:
P (n) = p(1 – p)n – 1
P (n) = 0.80(0.20)n – 1
P (2) = 0.80(0.20)1
P (2) = 0.16
There is a 16% chance that the first time you hit the target will be on your second attempt.

11. Example–Geometric Probability
Find the probability you will hit the target for the first time before your fifth attempt.
Solution:
P (n) = 0.80(0.20)n – 1
P (1) = 0.80(0.20)0 =0.80
P (2) = 0.80(0.20)1 =0.16
P (3) = 0.80(0.20)2 =0.032
P (4) = 0.80(0.20)3 =0.0064
P (1) + P (2) + P (3) +P (4) =
There is a 99.84% chance that the first time you hit the target will be before your fifth attempt.

12. Example–Geometric Probability
Find the probability you will hit the target for the first time on your fifth attempt or after.
Solution:
P (n ≥ 5)
P (n ≥ 5) = 1- P (n < 4)
P (n < 4) =
P (n ≥ 5) = =
There is a 0.16% chance that the first time you hit the target will be on your fifth attempt or after.

13. Example–Geometric Probability
A missile has a failure probability of 2%. Find the probability the first failure will happen on the twentieth attempt.
Solution:
P (n) = p(1 – p)n – 1
P (n) = 0.02(0.98)n – 1
P (20) = 0.02(0.98)20-1
P (20) = 0.02(0.98)19
P (2) =
There is a 1.36% chance that the first missile failure will happen on the twentieth attempt.

14. 26 The Geometric Probability Distributions
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