7-3 Part I: Conditional Probability,
Intersection of Events
CONDITIONAL EVENTS
Experiment: I flip a coin twice
E1 = two heads = {HH}
P(E1) = ¼
BUT…
suppose I tell you that the outcome was the same for both
flips, that is, that event
E2 = {HH, TT} has occurred
How would you assess the probability that E1 occurred
GIVEN THAT E2 has occurred?
written: P(E1 GIVEN E2) or P(E1 | E2)
S = Sample Space
E1
HH
TT
E2
HT
TH
Note: since we know that E2 has occurred, our sample
space has been reduced to just by that assumption:
sample space is now E2 = {TT, HH}
and thus P(E1 | E2) = 1 out of 2 = ½
7-3 Part I
p. 1
In general:
P(E1 GIVEN E2) = P(E1 | E2) =
=
n (E1  E 2)
n (E 2)
n (E1  E 2) / n (S) P(E1  E 2) 1 4
=
=
=½
12
n (E 2) / n (S)
P(E 2)
P(A GIVEN B)
For any two events A and B
P(A | B) =
P ( A  B)
P ( B)
THE AND (INTERSECTION) OF TWO EVENTS
Random experiment: flip coin twice
Event 1: E1 = get head on first throw = {HH, HT}
Event 2: E2 = get the same on both throws = {HH, TT}
Event E3: head on the 1st throw AND same on both throws
P(E 2  E1)
P(E1)
So: P(E1  E2) = P(E1)P(E2 | E1) = (1/2)(1/2) = ¼
We already know:
P(E2 | E1) =
P(A AND B)
For any two events A and B
P(A and B) = P(A  B) = P(A)P(B | A)
7-3 Part I
p. 2
Probability Trees
A box contains 3 blue and 2 white balls. Two balls are
drawn in succession, without replacement. Find the
probability of drawing a white ball on the second draw.
Event notation: B1 = blue on first draw
W2 = white on second draw, etc.
Way 1: P(W2) = P(W1)P(W2|W1) + P(B1)P(W2|B1)
=
(2/5) (1/4)
+ (3/5) (2/4)
= 8/20 = 2/5
Way 2: first draw a tree that completely analyzes the
problem
P(W1W2)
P(W2 | W1)
P(W1)
2/5
1/4
W1
W2
2/51/4 = 1/10
3/4
B2 P(W1B2)=3/10
Start
1/2
3/5
W2
B1
3/10
1/2
B2
3/10
So P(W2) = 1/10 + 3/10 = 4/10 = 2/5
Can also get P(B2) = 3/10 + 3/10 = 6/10 = 3/5 (naturally!)
7-3 Part I
p. 3
Probability tables
 probabilities can be displayed in a standard tabular form
 often starting as tables of counts
 see the following classification of 200 adults
 according to gender and educational attainment
 expressed as an empirical probability table ( 200):
Education
Elementary
Secondary
College
Totals
Male
.11
.22
.11
.44
Female
.195
.28
.085
.56
Totals
.305
.5
.195
1
Vocabulary:
 the darkly shaded cells are called joint probabilities
 another name for the probability of the AND of two events
 the lightly shaded cells are called marginal probabilities
(because they are shown in the margins of the table)
 the marginal probabilities represent certain totals of joint
probabilities, as shown.
Compute some probabilities (C=College F=Female M=Male):
P(C) = .195
P(C | F) = P(C  F)/P(F) = .085/.56 = .15
P(C | M) = P(C  M)/P(M) = .11/.44 = .25
Notice that a conditional probability is:
(a joint probability)  (a marginal probability)
7-3 Part I
p. 4