Topics in Mathematics 201-BNJ-05
Vincent Carrier
Conditional Probability
The conditional probability of an event A given that another event B occured is
P (A|B) =
P (A ∩ B)
.
P (B)
This definition can be understood in terms of area in a Venn diagram and by noticing
that if B occured, then B replaces Ω as the set of all possible outcomes.
Ω
A
.
....
.....
......
.......
........
.........
..........
.........
..........
.........
..........
.........
..........
.........
........
.......
........
.....
....
...
..
B
Example: A person in chosen at random in a certain country. Let
A : person has blue eyes
B : person is blond
and assume that
P (A) = 0.4,
P (B) = 0.3,
P (A ∪ B) = 0.45,
P (A ∩ B) = 0.25.
a) What is the probability that she is blond given that she has blue eyes?
P (B|A) =
0.25
P (B ∩ A)
=
= 0.625.
P (A)
0.4
b) What is the probability that she does not have blue eyes given that she is blond?
P (AC |B) = 1 − P (A|B) = 1 −
P (A ∩ B)
0.25
=1−
' 0.167.
P (B)
0.3
c) What is the probability that she is blond, given that she does not have blue eyes?
P (A|B C ) =
P (A ∩ B C )
P (A \ B)
P (A ∪ B) − P (B)
0.45 − 0.3
=
=
=
' 0.214.
P (B C )
P (B C )
1 − P (B)
1 − 0.3
The definition of conditional probability means that
P (A ∩ B) = P (A|B) P (B) = P (B|A) P (A).
Consider a sequence of events in time A1 , A2 , A3 , A4 . . . Then
P (A2 ∩ A1 ) = P (A2 |A1 ) P (A1 )
P (A3 ∩ A2 ∩ A1 ) = P (A3 |A2 ∩ A1 ) P (A2 ∩ A1 )
= P (A3 |A2 ∩ A1 ) P (A2 |A1 ) P (A1 )
P (A4 ∩ A3 ∩ A2 ∩ A1 ) = P (A4 |A3 ∩ A2 ∩ A1 ) P (A3 ∩ A2 ∩ A1 )
= P (A4 |A3 ∩ A2 ∩ A1 ) P (A3 |A2 ∩ A1 )P (A2 |A1 ) P (A1 )
In general,
P (An ) = P (A1 ) P (A2 |A1 ) P (A3 |A1 ∩ A2 ) · · · P (An |A1 ∩ A2 ∩ · · · ∩ An−1 ).
Example: An urn contains
12 green balls
and
8 yellow balls.
Seven balls are picked without replacement.
a) What is the probability of obtaining GYGGYGY in that order?
12 8 11 10 7 9 6
P (GYGGYGY) =
·
·
·
·
·
· .
20 19 18 17 16 15 14
b) What is the probability of obtaining YYYGGGG in that order?
8 7 6 12 11 10 9
·
·
·
·
·
·
= P (GYGGYGY).
P (YYYGGGG) =
20 19 18 17 16 15 14
c) What is the probability of obtaining 4 green balls and 3 yellow balls?
12 8
4
3
P (4G, 3Y) = .
20
7
d) What is the probability of obtaining GYGGYGY in that order (bis)?
12 8
12 8
7!
4! 3!
4
3
4
3
÷
·
P (GYGGYGY) =
=
.
20
20
4! 3!
7!
7
7