Download Lecture 7 Conditional probability

Conditional probability
Ordinary Probability
You are dealt two cards from a deck. What
is the probability the second card dealt is a
Jack?
We reason that if the two cards have been
delt, the probabiity that the first is a jacek
and the probabiity that the second is a jack
are identical. Since there are 4 jacks in the
deck, we compute
P(second is jack) = 4/52 = 1/13.
Conditional Probability
You are dealt two cards from a deck. What
is the probability of the second card was a
jack given the first card was not a jack is
called card dealt is a Jack?
Conditional Probability
You are dealt two cards from a deck. What
is the probability that the second card was a
jack given the first card was not a jack.
Conditional Probability
You are dealt two cards from a deck. What
is the probability of the second card was a
jack given the first card was not a jack is
called card dealt is a Jack?
4
Answer :
51
Reasoning
Once the first card has been delt, there are
51 cards remaining, and, since the first was
NOT a jack, there are 4 jacks in the set of 51
cards.
Conditional Probability
The probability of drawing jack given the first
card was not a jack is called conditional
probability. A key words to look for is
“given that.”
We will use the notation:
P(second a jack | first not a jack) = 4/51
General Conditional
Probability
The probability that the event A occurs,
given that B occurs is denoted:
This is read the probability of A given B.
P ( A | B ).
Conditional Probability
How would we draw the event A given B?
A
B
A
and
B
Conditional Probability
How would we draw the event A given B?
Since we know B has occurred, we ignore
everything else.
A
B
A
and
B
Conditional Probability
How would we draw the event A given B?
Since we know B has occurred, we ignore
everything else.
A
B
A
and
B
Conditional Probability
How would we draw the event A given B?
Since we know B has occurred, we ignore
everything else.
B
A
and
B
With some thought this tells us:
P( A and B)
P( A | B) 
P( B)
Additional notes
In the case of a equi-probability space,
we can reason that, since we know the
outcome is in B, we can use the set B as
our reduced sample space. The
probability P(A|B) can then be computed
as the number of points in A∩B as a
fraction of the number of points in B.
Continuing…
This gives
𝑛(𝐴 ∩ 𝐵)
𝑃 𝐴𝐵 =
𝑛(𝐵)
Dividing top and bottom by the numbers of
points in the original sample space S:
𝑛(𝐴 ∩ 𝐵)/𝑛(𝑆) 𝑃(𝐴 ∩ 𝐵)
𝑃 𝐴|𝐵 =
=
𝑛(𝐵)/𝑛(𝑆)
𝑃(𝐵)
Example from well
contamination
Private
Public
Bars show percents
Percent
75%
50%
25%
79%
21%
60%
40%
Below Limit
Detect
Below Limit
Detect
0%
MTBE-Detect
MTBE-Detect
Of private wells, 21% are
contaminated. Therefore
P(C|Private)=0.21
Of public wells, 40% are
contaminated. Therefore
P(C|Public)=0.40
Law of conditional probability
𝑃(𝐴 ∩ 𝐵)
𝑃 𝐴𝐵 =
𝑃(𝐵)
Multiplication Rule
𝑃 𝐴 = 𝑃 𝐵 𝑃(𝐴|𝐵)
Example
A local union has 8 members, 2 of whom are
women. Two are chosen by a lottery to
represent the union.
Example
A local union has 8 members, 2 of whom are
women. Two are chosen by a lottery to
represent the union.
a) What is the probability they are both
male?
Example
A local union has 8 members, 2 of whom are
women. Two are chosen by a lottery to
represent the union.
a) What is the probability they are both
male?
Answer : P(M)  P(M | M)
Example
A local union has 8 members, 2 of whom are
women. Two are chosen by a lottery to
represent the union.
a) What is the probability they are both
male?
6 5
Answer :   0.536
8 7
Example
A local union has 8 members, 2 of whom are
women. Two are chosen by a lottery to
represent the union.
b) What is the probability they are both
female?
Example
A local union has 8 members, 2 of whom are
women. Two are chosen by a lottery to
represent the union.
b) What is the probability they are both
female?
2 1
Answer : P(F) * P(F | F)    0.036
8 7
Example
Find the probability of selecting an all male
jury from a group of 30 jurors, 21 of whom
are men.
Example
Find the probability of selecting an all male
jury from a group of 30 jurors, 21 of whom
are men.
Solution:
P(12 M)
=P(M)*P(M|M)*P(M|MM) * ….
= 21/30 * 20/29 * 19/28 * 18/27 * … 10/19
= 0.00340
Independent Events
Two events A and B are independent if the
occurrence of one does not affect the
probability of the other.
Independent Events
Two events A and B are independent if the
occurrence of one does not affect the
probability of the other.
Two events A and B are independent then
P(A|B) = P(A).
Independent Events
Two events A and B are independent if the
occurrence of one does not affect the
probability of the other.
Two events A and B are independent then
P(A|B) = P(A).
Two events which are not independent are
dependent.
Multiplication Rule
Multiplication Rule:
For any pair of events:
P(A and B) = P(A) * P(B|A)
Multiplication Rule
Multiplication Rule:
For any pair of events:
P(A and B) = P(A) * P(B|A)
For any pair of independent events:
P(A and B) = P(A) * P(B)
Multiplication Rule
For any pair of events:
P(A and B) = P(A) * P(B|A)
For any pair of independent events:
P(A and B) = P(A) * P(B)
If P(A and B) = P(A) * P(B), then A and B
are independent.
Multiplication Rule
Multiplication Rule:
P(A and B) = P(A) * P(B) if A and B are
independent.
P(A and B) = P(B) * P(A|B) if A and B are
dependent.
Note: The multiplication rule extends to several events:
P(A and B and C) =P(C)*P(B|C)*P(A|BC)
Example
A study of 24 mice has classified the mice by two
categories
Fur Colour
Black White Grey
Eye
Colour
Red Eyes
3
5
2
Black Eyes
1
7
6
A study of 24 mice has classified the mice by two
categories
Fur Colour
Black White Grey
Eye
Colour
Red Eyes
3
5
2
Black Eyes
1
7
6
a) What is the probability that a randomly selected
mouse has white fur?
b) What is the probability it has black eyes given
that it has black fur?
c) Find pairs of mutually exclusive and
independent events.
A study of 24 mice has classified the mice by two
categories
Fur Colour
Black White Grey
Eye
Colour
Red Eyes
3
5
2
Black Eyes
1
7
6
a) What is the probability that a randomly selected
mouse has white fur?
12/24=0.5
b) What is the probability it has black eyes given
that it has black fur?
c) Find pairs of mutually exclusive and
independent events.
A study of 24 mice has classified the mice by two
categories
Fur Colour
Black White Grey
Eye
Colour
Red Eyes
3
5
2
Black Eyes
1
7
6
a) What is the probability that a randomly selected
mouse has white fur?
12/24=0.5
b) What is the probability it has black eyes given
that it has black fur?
1/4=0.25
c) Find pairs of independent events.
Fur Colour
Black White Grey
Eye
Colour
Red Eyes
3
5
2
Black Eyes
1
7
6
b) What is the probability it has black eyes given
that it has black fur?
1/4=0.25
c) Find pairs of mutually exclusive and
independent events.
IND: White Fur and Red Eyes;
Black Fur and Red Eyes
Descriptive Phrases
Descriptive Phrases require special care!
–
–
–
–
At most
At least
No more than
No less than
Review
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Conditional Probabilities
Independent events
Multiplication Rule
Tree Diagrams