Section 3.5 – Conditional Probability
Experiment:
Toss a single (balanced) die. Observe the up face.
S={1, 2, 3, 4, 5, 6}
P(1) = P(2) =…P(6) = 1/6
Let
A = {Toss an even #} = {2, 4, 6}
B = {Toss a #  3} = {1, 2, 3}
4
A
1
B
2
6
3
5
S
What is the probability of observing an even number?
P(Toss an even #) = P(A) = P(1) + P(2) + P(3) = 3/6 = 1/2
Does the probability of observing an even number change if you know that the # was  3, i.e. that event B
occurred?
Out of the 3 (equally likely) sample points in the event B, that was known to have occurred, only one sample
point, i.e. 2, was even, so…
P(even # given that the # was  3) = # evensample points 3 =1/3
# sample points 3
The condition “you know that the # was  3” causes the event B to become the reduced sample space of a
new conditional experiment.
def: The conditional probability of an event A is the probability that event A occurs knowing that the event
B has occurred on a single replication of the experiment.
Notation: P(A|B) – “the probability of A given B”
Above, we calculated
P(A|B) = # evensample points 3 = 1/3
# sample points 3
This only works if the sample points are equally likely.
4
A
General formula for a conditional probability
P( A / B)
P( A B)
P(B)
3.5 Conditional Probability - 1
B
2
6
5
1
3
S
Example: 1000 people classified by smoker & presence of cancer.
Yes
No
Smoker
Let
Yes
50
30
80
Cancer
No
200
720
920
250
750
1000
C
S = { smoker}
Sc = {non-smoker}
C = { cancer}
Cc = { no cancer}
P(S) =
P(Sc) =
P(C ) =
P(Cc ) =
P(C|S) = P(C

S)/P(S) =
P(C|Sc) = P(C

Sc)/P(Sc) =
S
You have a 5 times greater risk of getting cancer if you smoke.
What are the chances you are a smoker (non-smoker) if you have cancer?
P(S|C) =
P(Sc|C) =
P(S) = .25; P(S’) = .75; P(C) = .08; P(C’) = .92; P(C|S) = .2; P(C|S’) = .04; P(S|C) = 5/8; P(S’|C) = 3/8
3.5 Conditional Probability - 2