WARM-UP
9/4/12
Fifteen persons reporting to a Red Cross center one day are
typed for blood, and the following counts are found:
Blood group
O
A
B
AB
Total
No. of
persons
3
5
6
1
15
If one person is randomly selected, what is the probability that
this person’s blood group is:
a) AB?
b) Either A or B?
c) Not O?
ADDITION LAW
P(AUB)=P(A)+P(B)-P(AB)
Special Addition Law for Incompatible Events
P(AUB)=P(A)+P(B)
QUESTION: 52
• What is the probability of getting a card that is
either a black card or and ace?
DEMONSTRATION
Conditional Probability
The Basics
QUESTION 1:
• Suppose we have 14 marbles in a bag; 6 are
green, 4 are red and 4 are yellow. What is the
probability of pulling a red marble, given that
the marble we just pulled is a primary color?
14 MARBLES
• 6 Green, 4 Red, 4 Yellow
14 MARBLES
• P(A)= probability of getting a red marble
• P(B)= probability of getting a marble that’s a
primary color
P(B)=8/14
• 8 of the 14 marbles are primary colors
P(AB)=4/14
• 4 of the 8 marbles are red
P(A|B)=P(AB)/P(B)=4/8
• We were told marble was primary color, and
only 4 of those marbles could possibly be red
P(A|B)=P(AB)/P(B)
• The probability of an event A, given that an
event B has already occurred
• How likely will A happen, since B has already
happened?
QUESTION:
• A family has two children. What is the
conditional probability that both are boys
given that at least one of them is a boy?
SAMPLE SPACE
P(A)=getting at least one boy
• 3 of 4 outcomes gives at least one boy
P(A)=getting at least one boy
• 1 out of the 3 outcomes gives us both boys
P(A)=getting at least one boy
P(B|A)=P(AB)/P(A)=1/3
BODY WEIGHT AND HYPERTENSION
Overweight Normal Weight
Underweight
Total
Hypertensive
.10
.08
.02
.20
Not
Hypertensive
.15
.45
.20
.80
Total
.25
.53
.22
1.00
a) What is the probability that a person selected at random
from this group will have hypertension?
b) A person selected at random from this group is found to be
underweight. What is the probability that this person is also
hypertensive?
BODY WEIGHT AND HYPERTENSION
Overweight Normal Weight
Underweight
Total
Hypertensive
.10
.08
.02
.20
Not
Hypertensive
.15
.45
.20
.80
Total
.25
.53
.22
1.00
a) Because 20% of the group is hypertensive and the individual
is selected at random from this group, we conclude that
P(A)=.2. This is unconditional probability of A
BODY WEIGHT AND HYPERTENSION
Overweight Normal Weight
Underweight
Total
Hypertensive
.10
.08
.02
.20
Not
Hypertensive
.15
.45
.20
.80
Total
.25
.53
.22
1.00
b) When given the info that the selected person is underweight,
we ignore the first and second columns. The third column
shows that among the subgroup of underweight people the
proportion of having hypertension is .02/.22=.091.
KEY FORMULAS
CONDTIONAL PROBABILITY
P(A|B)=P(AB)/P(B)
MULTIPLICATION LAW OF PROBABILITY
P(AB)=P(B)P(A|B)
COMPREHENSION EXAMPLE
• http://www.khanacademy.org/math/probabili
ty/v/independent-events-1
Let’s see if you understand what it means for
two events to be independent or dependent
UNDERSTANDING INDEPENDENCE
There are 25 pens in a container on your desk.
Among them, 20 will write well but 5 have
defective ink cartridges. You will select 2 pens
to take a business appointment. What is the
probability that both pens are defective?
• Without replacement
• With replacement
INDEPENDENCE RELATION
• Two events A and B are independent if
P(A|B)=P(A)
Equivalently
P(B|A)=P(B)
Or
P(AB)=P(A)P(B)
APPLICATION
• http://www.khanacademy.org/math/probabili
ty/v/independent-events-2
In this problem, we are using the formula:
P(AB)=P(A|B)P(B)
Since the two events are independent
P(AB)=P(A)P(B), where P(A|B)=P(A)
DRILL SET
• Now, let’s practice what you have learned so
far. You will have 2 to 3 minutes to solve each
problem; I will be keeping track of time.