Download Sect. 3-2 Conditional Probability

Section 3.2
Conditional Probability
and the
Multiplication Rule
Conditional Probability
The probability an event B will occur, given (on the
condition) that another event A has occurred.
We write this as P(B|A) and say “probability of B, given A.”
Two cars are selected from a production line of 12
cars where 5 are defective. What is the probability the
2nd car is defective, given the first car was defective?
Given a defective car has been selected, the conditional
sample space has 4 defective out of 11. P(B|A) = 4/11
Independent Events
Two dice are rolled. Find the probability
the second die is a 4 given the first was a 4.
Original sample space: {1, 2, 3, 4, 5, 6}
Given the first die was a 4, the conditional
sample space is: {1, 2, 3, 4, 5, 6}
The conditional probability, P(B|A) = 1/6
Independent Events
Two events A and B are independent if the
probability of the occurrence of event B is not
affected by the occurrence
(or non-occurrence) of event A.
A = Being female
B = Having type O blood
A = 1st child is a boy
B = 2nd child is a boy
Two events that are not independent are dependent.
A = taking an aspirin each day
B = having a heart attack
A = being a female
B = being under 64” tall
Independent Events
If events A and B are independent, then P(B|A) = P(B)
Conditional Probability
Probability
12 cars are on a production line where 5 are defective and 2 cars are
selected at random.
A = first car is defective
B = second car is defective.
The probability of getting a defective car for the second car depends
on whether the first was defective. The events are dependent.
Two dice are rolled. A = first is a 4 and B = second is a 4
P(B) = 1/6 and P(B|A) = 1/6. The events are independent.
Contingency Table
The results of responses when a sample of adults in 3
cities was asked if they liked a new juice is:
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
One of the responses is selected at random. Find:
1. P(Yes)
2. P(Seattle)
3. P(Miami)
4. P(No, given Miami)
Total
400
350
250
1000
Solutions
Yes
No
Undecided
Total
Omaha
100
125
75
300
1. P(Yes)
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
= 400 / 1000 = 0.4
2. P(Seattle)
= 450 / 1000 = 0.45
3. P(Miami)
= 250 / 1000 = 0.25
4. P(No, given Miami)
= 95 / 250 = 0.38
Answers: 1) 0.4 2) 0.45 3) 0.25 4) 0.38
Multiplication Rule
To find the probability that two events, A and B will occur in
sequence, multiply the probability A occurs by the conditional
probability B occurs, given A has occurred.
P(A and B) = P(A) x P(B|A)
Two cars are selected from a production line of 12 where 5 are
defective. Find the probability both cars are defective.
A = first car is defective B = second car is defective.
P(A) = 5/12
P(B|A) = 4/11
P(A and B) = 5/12 x 4/11 = 5/33 = 0.1515
Multiplication Rule
Two dice are rolled. Find the probability both are 4’s.
A = first die is a 4 and B = second die is a 4.
P(A) = 1/6
P(B|A) = 1/6
P(A and B) = 1/6 x 1/6 = 1/36 = 0.028
When two events A and B are independent, then
P (A and B) = P(A) x P(B)
Note for independent events P(B) and P(B|A) are the
same.
Homework 1-13 all pgs.135-136
Day 2: 14- 26 all pgs.136-138