Lecture 7
Tree Diagrams
Slide 1
1st question 2nd question
The tree diagram (right) summarizes
the possible outcomes for a
true/false question followed
by a multiple choice question with
five possible answers (a, b, c, d, e).
Note that there are 10
possible combinations.
Example
Slide 2
Genetics Experiment Mendel’s famous hybridization
experiments involved peas, like those shown below. If
two of the peas shown in the figure are randomly selected
without replacement, find the probability that the first
selection has a green pod and the second has a yellow
pod.
8 green pod
6 yellow pod
Example - Solution
Slide 3
First selection: P(1st green pod) = 8/14
(14 peas, 8 of which have green pods)
Second selection:
P(2nd yellow pod given 1st green pod) = 6/13
(13 peas remaining, 6 of which have yellow pods)
P( First pea with green pod and second pea with yellow pod)
=
8
6
≈ 0.264
•
14 13
= P(1st green pod) • P(2nd yellow pod given 1st green pod)
Example
Important Principle
Slide 4
The preceding example illustrates the
important principle that the probability for the
second event B should take into account the
fact that the first event A has already occurred.
Slide 5
Conditional Probability
Let A and B be two events.
P(B|A) represents the probability of event B
occurring after it is assumed that event A has
already occurred (read B|A as “B given A.”)
Multiplication Rule:
Basic case
Let
Slide 6
A be an event from first trial
B be an event from second trial
P(A and B) = P(A) • P(B|A)
(Multiplication Rule)
Conditional Probability
General case
Slide 7
P(B A) represents the probability of event B
occurring after it is assumed that event A has
already occurred (read B A as “B given A.”)
P(B A) =
P(A and B)
P(A)
Slide 8
Independent Events
Two events A and B are independent if the
occurrence of one does not affect the
occurrence of the other.
P(B|A) = P(B)
If A and B are not independent, they are said to
be dependent.
Multiplication Rule
Slide 9
! P(A and B) = P(A) • P(B A)
! Note that if A and B are independent
events, P(B A) = P(B). Then
P(A and B)=P(A) • P(B)
Small Samples from
Large Populations
Slide 10
If a sample size is no more than 5% of
the size of the population, treat the
selections as being independent (even
if the selections are made without
replacement, so they are technically
dependent).
Example
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Slide 11
Girls
27
18
56
Totals
706
1517
2223
If one of the 2223 people is randomly selected,
find the probability that the person is a boy
given that the person survived.
29
P(boy | survivied) =
= 0.79
706
Example
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Slide 12
Girls
27
18
56
Totals
706
1517
2223
If 1 of 2223 people is randomly selected, find the
probability that the person died given that the
person is a man.
1360
P(died | man) =
= 0.80
1692
Testing for Independence
Slide 13
Recall: Events A and B are independent if the
occurrence of one does not affect the probability of
occurrence of the other. This suggests the
following test for independence:
Two events A and B are
independent if
Two events A and B are
dependent if
P(B A) = P(B)
P(B A) = P(B)
or
or
P(A and B) = P(A) P(B)
P(A and B) = P(A) P(B)
Example
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Slide 14
Girls
27
18
56
Totals
706
1517
2223
If 1 of 2223 people is randomly selected, find the
probability that the person is died.
1517
P(died) =
= 0.68 ≠ P(died | man )
2223
Thus, the events {died} and {man} are dependent.
Using complement event
Slide 15
Key words of understanding
! “At least one” is equivalent to “one or
more.”
! The complement of getting at least one item
of a particular type is that you get no items of
that type.
Example
Slide 16
Gender of Children Find the probability of a couple
having at least 1 girl among 3 children. Assume that
boys and girls are equally likely and that the gender of
a child is independent of the gender of any brothers or
sisters.
Solution
Step 1: Let A = {at least 1 of the 3 children is a girl}
Step 2: Ā ={not child among 3 children is girl}
= {all 3 children are boys}
= {boy and boy and boy}
Solution (cont’d)
Example
Step 3: Find the probability of the complement.
P(Ā) = P(boy and boy and boy)
= P(boy) • P(boy) • P(boy)
1 1 1 1
= • • =
2 2 2 8
Step 4: Find P(A) by evaluating 1 – P(Ā).
1 7
P ( A) = 1 − P ( A) = 1 − =
8 8
Interpretation There is a 7/8 probability that if a
couple has 3 children, at least 1 of them is a girl.
Slide 17
Key Principle
Slide 18
To find the probability of at least one of
something, calculate the probability of
none, then subtract that result from 1.
That is,
P(at least one) = 1 – P(none)