Winter 2013 STAT 302
Chapter 3 Conditional Probability and Independence
In-class activity
The odds that a randomly chosen individual has a circulation problem is
0.25. Individuals who have a circulation problem are three times as likely
to be smokers as those who do not have a circulation problem. Given
that an individual is a smoker, what is the probability that he/she has a
circulation problem?
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
1. What is the probability that a randomly chosen individual
has a circulation problem?
A.
B.
C.
D.
E.
0.10
0.20
0.25
0.33
0.50
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
2. Suppose we define the following events:
C = an individual has a circulation problem
T = an individual is a smoker
Which of the following is correct?
A.
B.
C.
D.
E.
P (T ) = 3P (T c)
P (C|T ) = 3P (C|T c)
P (T |C) = 3P (T |C c)
P (T ∩ C) = 3P (T c ∩ C)
P (T ∩ C) = 3P (T ∩ C c)
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
3. Given that an individual is a smoker, what is the probability
that he/she has a circulation problem?
A.
B.
C.
D.
E.
3/7 = 0.4285
3/4 = 0.75
1/13 = 0.0769
2/3 = 0.6667
5/9 = 0.5556
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
Let’s look at the solution .....
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
We define the following events:
C = an individual has a circulation problem
T = an individual is a smoker
P (C) =
0.25
1+0.25
= 0.20
P (T |C) = 3P (T |C c)
P (C|T ) =
=
=
=
P (T |C)P (C)
P (T |C)P (C) + P (T |C c)P (C c)
P (T |C)P (C)
P (T |C)P (C) + (1/3)P (T |C)[1 − P (C)]
P (C)
P (C) + (1/3)[1 − P (C)]
0.20
0.20 + (1/3)[1 − 0.20]
= 3/7
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
Conditional probability
Given that an event F has occurred [P (F ) > 0], the probability that
another event E occurs is called the conditional probability of E given F .
Notation and formula:
P (E ∩ F )
P (E|F ) =
P (F )
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
Note:
1. P (E|F ) 6= P (F |E)
2. P (E c ∩ F ) = P (F ) − P (E ∩ F )
3. P (E c|F ) = 1 − P (E|F )
Proof : P (E c|F ) =
=
P (E c ∩ F )
P (F )
P (F ) − P (E ∩ F )
P (F )
P (E ∩ F )
= 1−
P (F )
= 1 − P (E|F )
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
Properties of P (E|F )
1. 0 ≤ P (E|F ) ≤ 1
Proof: P (E|F ) =
P (E∩F )
P (F )
Here P (E ∩ F ) ≥ 0 and P (F ) > 0, so P (E|F ) ≥ 0
Also, {E ∩ F } ⊂ F ⇒ P (E ∩ F ) ≤ P (F ) , so P (E|F ) ≤ 1
2. P (F |F ) = 1; P (S|F ) = 1
Proof : P (F |F ) =
P (F ∩ F ) P (F )
=
=1
P (F )
P (F )
P (S|F ) =
P (S ∩ F ) P (F )
=
=1
P (F )
P (F )
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
3. For a sequence of mutually exclusive events E1, E2, ...(Ei ∩ Ej = φ
for all i 6= j),
∞
∞
[
X
P
Ei|F =
P (Ei|F )
i=1
Proof : P
∞
[
Ei|F
=
i=1
P
S∞
i=1 Ei
S∞
i=1 (Ei ∩ F )
P (F )
P
P∞
=
∩F
P (F )
i=1
=
i=1 P (Ei
by the Distributive law
∩ F)
P (F )
since {Ei ∩ F }0s are mutually exclusive
∞
X
=
P (Ei|F )
i=1
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
Bayes’ formula
Total probability: P (E) =
Pn
i=1 P (E|Fi )P (Fi )
The Bayes’ formula:
P (Fi|E) =
P (Fi ∩ E)
P (E|Fi)P (Fi)
= Pn
P (E)
j=1 P (E|Fj )P (Fj )
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
Odds of an event
• The odds of an event E is defined as the ratio of the probability of an
event E to the probability of its complement
P (E)
P (E)
=
P (E c) 1 − P (E)
• The odds tells how much more likely E occurs compared to its
complement.
– if odds > 1, E is more likely than E c
– if odds < 1, E is less likely than E c
– if odds = 1, E and E c are equally likely
Eugenia Yu, UBC Department of Statistics
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Winter 2013 STAT 302
• Suppose the odds of E is α, then P (E) = αP (E c)
P (E) + P (E c) = 1
αP (E c) + P (E c) = 1
P (E c) =
P (E) =
Eugenia Yu, UBC Department of Statistics
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1+α
α
1+α
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