Download Section 3

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Pattern recognition wikipedia , lookup

Randomness wikipedia , lookup

Simulated annealing wikipedia , lookup

Least squares wikipedia , lookup

Probability box wikipedia , lookup

Multiplication algorithm wikipedia , lookup

Probabilistic context-free grammar wikipedia , lookup

Birthday problem wikipedia , lookup

Probability amplitude wikipedia , lookup

Transcript
Lesson 5 - 3
Independence and the
Multiplication Rule
Objectives
• Understand independence
• Use the Multiplication Rule for independent events
• Compute at-least probabilities
Vocabulary
• Independent – two events are independent if the
occurrence of E does not affect the probability of
event F
• Dependent – two events are dependent if the
occurrence of E affects the probability of event F
Multiplication Rules
for Independent Events
If E and F are independent events,
then P(E and F) = P(E) ∙ P(F)
If events E, F, G, ….. are independent, then
P(E and F and G and …..) = P(E) ∙ P(F) ∙ P(G) ∙ ……
Example: P(rolling 2 sixes in a row) = ??
Example: P(rolling 5 sixes in a row) = ??
At least Probabilities
P(at least one) = 1 – P(complement of “at least one”)
= 1 – P(none)
Example: P(rolling a least one six in three rolls) = ??
= 1 - P(none)
= 1 – (5/6)• (5/6)• (5/6)
= 1 – 0.5787 = 0.4213
Example 1
There are two traffic lights on the route used by Pikup Andropov
to go from home to work. Let E denote the event that Pikup must
stop at the first light and F in a similar manner for the second
light. Suppose that P(E) = .4 and P(F) = .3 and P(E and F) = .15.
What is the probability that he:
a) must stop for at least one light?
= 1 - P(none) = 1 – (0.6)• (0.7) = 1 – 0.42 = 0.58
b) doesn't stop at either light?
= (1-P(E)) • (1-P(F)) = 0.6 • 0.7 = 0.42
c) must stop at exactly one light?
= P(E) + P(F) – P(E and F) = 0.4 + 0.3 – 0.15 = 0.55
d) must stop just at the first light?
= 0.4
Example 2
Single-family
Condo
Multi-family
Adjustable
.40
.21
.09
Fixed
.10
.09
.11
A large lending institution gives both adjustable and fixed rate
mortgages on residential property. It breaks residential property into
three categories: single-family, condo, and multi-family. The
accompanying table, sometimes called a joint probability table
displays the probabilities for the problem:
a) What is the probability of having a fixed rate mortgage?
= 0.1 + 0.09 + 0.11 = 0.3
b) What is the probability of having an adjustable mortgage and
owning a condo?
= 0.21
c) What is the probability of having a fixed mortgage or owning a
single-family home?
= P(f) + P(sf) – P(f and sf) = 0.3 + 0.5 – 0.1 = 0.7
Summary and Homework
• Summary
– Disjoint events are not necessarily independent
– The Multiplication Rule applies to independent
events, the probabilities are multiplied to calculate an
“and” probability
– Probabilities obey many different rules
•
•
•
•
•
Probabilities must be between 0 and 1
The sum of the probabilities for all the outcomes must be 1
The Complement Rule
The Addition Rule (and the General Addition Rule)
The Multiplication Rule
• Homework
– pg 282 - 283: 11, 12, 17, 19, 23