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Independence and the Multiplication Rule
MATH 130, Elements of Statistics I
J. Robert Buchanan
Department of Mathematics
Fall 2015
Independent Events
Definition
Two events A and B are independent if the occurrence of event
A in a probability experiment does not affect the probability of
event B. Two events are dependent if the occurrence of event
A in a probability experiment affects the probability of B.
Examples
Are the following events independent or dependent?
1. Rolling a pair of dice and observing a 1 on the first die and
a 1 on the second die.
2. Drawing a heart from a regular deck of playing cards and
then drawing another heart without replacing the first card.
3. Drawing a heart from a regular deck of playing cards and
then drawing another heart after replacing the first card.
4. Owning a red automobile and having blond hair.
5. Owning a red automobile and having a flat tire today.
6. Studying for an exam and passing the exam.
7. Have a bachelor’s degree and earn more than $65,000 per
year.
Independence vs. Mutual Exclusivity
Remark: Mutually exclusive (or disjoint) events are dependent.
Why?
Multiplication Rule
Theorem (Multiplication Rule for Independent Events)
If A and B are independent events, then
P(A and B) = P(A) · P(B).
Examples
You draw a card from a regular deck of playing cards, replace it,
and draw another card from the deck.
1. What is the probability both cards were diamonds?
2. What is the probability the first card was a diamond and
the second card was an ace?
3. What is the probability both cards were aces?
N Independent Events
Theorem
If events A1 , A2 , . . . , AN are independent, then
P(A1 and A2 and . . . and AN ) = P(A1 ) · P(A2 ) · · · P(AN ).
Example
The probability that a randomly selected 24-year-old male will
survive the year is 0.9986 according to the National Vital
Statistics Report, Vol. 56, No. 9. What is the probability that 4
randomly selected 24-year-old males will survive the year?
Example
The probability that a randomly selected 24-year-old male will
survive the year is 0.9986 according to the National Vital
Statistics Report, Vol. 56, No. 9. What is the probability that 4
randomly selected 24-year-old males will survive the year?
P(all 4 survive) = P(1st and 2nd and 3rd and 4th survive)
= P(1st survives) · P(2nd survives)
· P(3rd survives) · P(4th survives)
= (0.9986)(0.9986)(0.9986)(0.9986)
= 0.9944
Example
What is the probability that 10 randomly selected people from
the general population are all male?
Example
What is the probability that 10 randomly selected people from
the general population are all male?
Let the probability of one individual being male be p = 1/2. In a
randomly selected group of ten individuals, the probability of
ten males is
10
1
P(10 males) =
= 0.000977.
2
“At least” Probabilities
P(at least one of something) = 1 − P(none of something)
“At least” Probabilities
P(at least one of something) = 1 − P(none of something)
Example
A fair coin is flipped eight times. What is the probability that
“tails” appeared at least once?
“At least” Probabilities
P(at least one of something) = 1 − P(none of something)
Example
A fair coin is flipped eight times. What is the probability that
“tails” appeared at least once?
P("tails" at least once) = 1 − P(no "tails")
= 1 − P(8 "heads")
8
1
= 1−
2
= 0.996
Example
Suppose that a satellite defense system is established in which
six satellites acting independently have a 0.99 probability of
detecting an incoming threat. What is the probability that at
least one of the six satellites will detect an incoming threat?
Would such a system make you feel safe?
Example
Suppose that a satellite defense system is established in which
six satellites acting independently have a 0.99 probability of
detecting an incoming threat. What is the probability that at
least one of the six satellites will detect an incoming threat?
Would such a system make you feel safe?
P(at least 1 detection) = 1 − P(no detections)
= 1 − (0.01)6
= 1 − 0.000000000001
= 0.999999999999
Birthday Problem (1 of 3)
Question: In a room of n ≥ 2 randomly selected people, what
is the probability at least two of them have the same birthday
(month and day, but not necessarily year)?
Birthday Problem (1 of 3)
Question: In a room of n ≥ 2 randomly selected people, what
is the probability at least two of them have the same birthday
(month and day, but not necessarily year)?
P(at least two same)
= 1 − P(all different)
364
363
362
365 − n + 1
= 1−
···
365
365
365
365
Birthday Problem (2 of 3)
n
10
15
20
30
32
35
40
50
60
P
0.117
0.253
0.411
0.706
0.753
0.814
0.891
0.970
0.994
Birthday Problem (3 of 3)
p
1.0
0.8
0.6
0.4
0.2
n
10
20
30
40
50
60
Summary
Remark: Probability statements containing “or” use the
Addition Rule, while statements containing “and” use the
Multiplication Rule.
Addition Rule: If A and B are disjoint,
P(A or B) = P(A) + P(B).
If A and B are not disjoint,
P(A or B) = P(A) + P(B) − P(A and B).
Complements: For any event E,
P(E c ) = 1 − P(E).
Multiplication Rule: If A and B are independent,
P(A and B) = P(A) · P(B).