Transcript C2_Math3033

MATH 3033 based on
Dekking et al. A Modern Introduction to Probability and Statistics. 2007
Slides by Tim Birbeck
Instructor Longin Jan Latecki
C2: Outcomes, events, and probability
2.1 – Sample Spaces
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Sample Space: A sample space is a set whose elements
describe the outcomes of the experiment in which we
are interested.
Example:
If we ask arbitrary people on the street what month they
were born, the following is an obvious sample space:
  {Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec}.
2.1 – Sample Spaces
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Permutation: the order in which n different objects can
be placed.
Example:
If we have three envelopes and number them 1, 2, and 3,
the following sample space consists of every different
permutation we can make using all three envelopes:
  {123, 132, 213, 231, 312, 321}
2.2 – Events
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Event: A subset of the sample space
Example:
In the birthday experiment, if we ask for the outcomes
that only involve the months with 31 days, we would have
the following event:
L  {Jan, Mar, May, Jul, Aug, Oct, Dec}.
2.2 – Events
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We can use set operators to combine events:
Name
Union
Intersection
Compliment
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Definition
C  A B
C  A B
C  Ac
C  {x : x  A  x  B}
C  {x : x  A  x  B}
C  {x : x  A}
Example:
In the birthday experiment, if we intersect the event R,
where the month has the letter ‘r’ in it, and the event L,
where the month has 31 days, we get the following:
L  R  {Jan, Mar, Oct, Dec}
2.2 – Events
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Disjoint / Mutually Exclusive: Two events that have no
outcomes in common. A∩B = ∅
Example:
In the birthday experiment, the event L, all the birthdays
with 31 days, and the event {Feb}
are mutually exclusive.
We say the event A implies event B if the outcomes of A
also lie in B. A ⊂ B
2.2 – Events
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DeMorgan’s laws: For any two events A and B we have:
(A ∪ B)c = Ac ∩ Bc and (A ∩ B)c = Ac ∪ B
2.3 – Probability
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Probability function: A probability function P on a
finite sample space Ω assigns to each event A in Ω a
number P(A) in [0,1] such that:
(i) P(Ω) = 1, and
(ii) P(A ∪ B) = P(A) + P(B) if A and B are disjoint.
The number P(A) is called the probability that A occurs.
Example: In an experiment where we flip a perfectly
weighted coin and record whether the coin lands on
heads or tails, we could define the probability function P
such that:
P({H}) = P({T}) =1/2
2.3 – Probability
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Formally, we should write P({T}) and not P(T) because a
probability function works on events and not outcomes.
However, in practice, we often drop the curly braces for a
singleton set.
If we consider an experiment that only has two
outcomes, such as success or failure, one outcome has a
probability p to occur where 0 < p < 1, and the other
outcome has a probability of 1 - p to occur.
2.3 – Probability
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To assign probability to an event, we can use the additivity
property.
Example:
Ω = {123, 132, 213, 231, 312, 321}
P(213) = P(231) = 1/6
T = {213, 231}
P(T) = P(213) + P(231) = 1/6 + 1/6 = 1/3
2.3 – Probability
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If two sets are not disjoint, we must use following rule to
determine probability:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Example:
Ω = {123, 132, 213, 231, 312, 321
P(213) = P(231) = P(123) =1/6
S = {123, 213}
T = {213, 231}
S ∩ T = {213}
P(S ∪T) = [P(123) + P(213)] + [P(213) + P(231)] - P(213)
= [1/3] + [1/3] – 1/6 = 1/2
2.4 – Products of sample spaces
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Usually, one experiment is not sufficient, so an
experiment is performed several times.
To get a sample space of multiple experiments, we use
the cross product.
Example:
Ω = Ω1 × Ω2 = {(ω1, ω2) : ω1 ∈ Ω1, ω2 ∈ Ω2}
where Ω1 is the sample space of the first experiment and
Ω2 is the sample space of the second experiment.
Example:
If we flip a coin twice the sample space would be:
Ω = {H, T}× {H, T} = {(H,H), (H, T), (T,H), (T,T)}.
2.4 – Products of sample spaces
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A certain experiment may have 2 outcomes: success or
failure. If we perform this experiment n times and let 0
represent failure and 1 represent success, we have the
following sample space:
Ω = Ω1 × Ω2 × … × Ωn
Where
Ω1 = Ω2 = … = Ωn = {1, 0}
2.4 – Products of sample spaces
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Example:
If a certain Dr. Latecki goes on 5 dates and the date is either
“successful” or “unsuccessful,” we can model the event where
Dr. Latecki is only successful on 1 of his 5 dates as:
A = {(0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (0, 0, 1, 0, 0), (0, 1, 0, 0, 0),
(1, 0, 0, 0, 0)}
Assuming Dr. Letecki’s chance of being “successful” is p and
each date’s chance of success is independent of the previous
date:
P(A) = 5 (1 − p)4 p1
Because: (1 – p) is the probability of being unsuccessful and this
must happen 4 times. p is probability of being successful and
this must happen 1 time. There are 5 outcomes in the event A.
2.5 – An infinite sample space
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A probability function on an infinite (or finite) sample
space Ω assigns to each event A in Ω a number P(A) in [0,
1] such that
(i) P(Ω) = 1, and
(ii) P(A1 ∪ A2 ∪ A3 ∪ ・ ・ ・) = P(A1) + P(A2) + P(A3) +
・・・
If A1,A2,A3, . . . are disjoint events.
2.5 – An infinite sample space
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Example:
If we flip a coin until it lands on heads, the outcome of the
experiment could be the number of times the coin
needed to be flipped until heads came up. The sample
space for this experiment would be:
Ω = {1, 2, 3, . . . }
If the chance of landing on heads is p, the chance of
landing on tails is 1-p. Therefore:
P(1) = p. P(2) = (1 – p)1p. P(3) = (1-p)2p
P(n) = (1-p)n-1p