1. Aim – How can we assess how one event’s outcome affects the outcome of another event?
H.W. – pg 333 – 334 #63 – 68 Do Now – Suppose you pick two cards, at random, from a standard deck of 52 cards that has been shuffled. If you pick one card, put it back, will this affect the outcome of the second card that is picked? How is the second outcome affected if you pick a card the first time, and do not put it back?

2. CHAPTER 5 Probability: What Are the Chances?
5.3
Conditional Probability and Independence

3. Conditional Probability and Independence
CALCULATE and INTERPRET conditional probabilities.
USE the general multiplication rule to CALCULATE probabilities.
USE tree diagrams to MODEL a chance process and CALCULATE probabilities involving two or more events.
DETERMINE if two events are independent.
When appropriate, USE the multiplication rule for independent events to COMPUTE probabilities.

4. What is Conditional Probability?
The probability we assign to an event can change if we know that some other event has occurred. This idea is the key to many applications of probability. When we are trying to find the probability that one event will happen under the condition that some other event is already known to have occurred, we are trying to determine a conditional probability.
The probability that one event happens given that another event is already known to have happened is called a conditional probability.
Suppose we know that event A has happened. Then the probability that event B happens given that event A has happened is denoted by P(B | A).
Read | as “given that” or “under the condition that”

5. Calculating Conditional Probabilities
To find the conditional probability P(A | B), use the formula
The conditional probability P(B | A) is given by

6. Calculating Conditional Probabilities
Consider the two-way table on page Define events
E: the grade comes from an EPS course, and
L: the grade is lower than a B.
Total
Total
6300
1600
2100
Find P(L)
Find P(E | L)
Find P(L | E)
Try Exercise 39
P(L) = 3656 / =
P(E | L) = 800 / 3656 =
P(L| E) = 800 / 1600 =

7. The General Multiplication Rule
The probability that events A and B both occur can be found using the general multiplication rule
P(A ∩ B) = P(A) • P(B | A)
where P(B | A) is the conditional probability that event B occurs given that event A has already occurred.
In words, this rule says that for both of two events to occur, first one must occur, and then given that the first event has occurred, the second must occur.

8. Example and practice:
The Pew Internet and American Life Project find that 93% of teenagers (ages 12 to 17) use the internet, and that 55% of online teens have posted a profile on a social – networking site.
Problem – Find the probability that a randomly selected teen uses the internet and has posted a profile. Show all work:
Solution:
P(online) = 0.93
P(profile|online) = 0.55
P(online and profile) = P(online) times P(profile|online)
P(online and profile) = (0.93)(0.55) =
There is about a 51% chance that a randomly selected teen used the internet and has posted a profile on a social – networking site.

9. Practice:
About 55% of high school students participate in a school athletic team at some level, and about 5% of these athletes go on to play on a college team in the NCAA.
Problem: What percent of high school students play a sport in high school and go on to play a sport in the NCAA?
Solution: We know P(high school sport) = 0.55 and P(NCAA sport | high school sport) = 0.05, so P(high school sport and NCAA sport) = P(high school sport) x P(NCAA sport | high school sport) = (0.55)(0.05) = Almost 3% of high school students will play a sport in high school and in the NCAA.

10. Exit Ticket:
A number is selected at random from the set {1, 2, 3, 4, 5, 6, 7, 8}
Find the probability of the following:
P(prime|odd)

11. Aim – How can tree diagrams be used to evaluate the probability of conditional events?
H.W. – pg. 334 – 335 #73 – 79 (odds only)

12. Tree Diagrams
The general multiplication rule is especially useful when a chance process involves a sequence of outcomes. In such cases, we can use a tree diagram to display the sample space.
Consider flipping a coin twice.
What is the probability of getting two heads?
Sample Space:
HH HT TH TT
So, P(two heads) = P(HH) = 1/4

13. Example: Tree Diagrams
The Pew Internet and American Life Project finds that 93% of teenagers (ages 12 to 17) use the Internet, and that 55% of online teens have posted a profile on a social-networking site.
What percent of teens are online and have posted a profile?
51.15% of teens are online and have posted a profile.

14. Example:
E-book Roger Federer:

15. Practice:
Shannon hits the snooze bar on her alarm clock on 60% of school days. If she doesn’t hit the snooze bar, there is a 0.90 probability that she makes it to class on time. However, if she hits the snooze bar, there is only a 0.70 probability that she makes it to class on time. On a randomly chosen day, what is the probability that Shannon is late for class? (Draw a tree diagram as well).
Here are the values:
P(late and hit the snooze) = (0.60)(0.30) = 0.18
P(late and did not hit the snooze) = (0.40)(0.10) = 0.04
P(late) = = 0.22 = 22%
Something to think about:

16. Practice continued:
Suppose that Shannon is late for school. What is the probability that she hit the snooze bar that morning?
Solution:
P(hit the snooze|late for school) = (0.18)/(0.22) = 0.818
Interpretation:
When Shannon is late, there is a probability that she hit the snooze bar that morning.

17. Practice (exit):
The Kaiser Family Foundation recently released a study about the influence of media in the lives of young people aged 8–18 ( In the study, 17% of the youth were classified as light media users, 62% were classified as moderate media users, and 21% were classified as heavy media users. Of the light users who responded, 74% described their grades as good (A’s and B’s), while only 68% of the moderate users and 52% of the heavy users described their grades as good. Suppose that we selected one young person at random.
(a) Draw a tree diagram to represent this situation.
(b) Find the probability that this person describes his or her grades as good.
(c) Given that this person describes his or her grades as good, what is the probability that he or she is a heavy user of media?

18. Aim – How can we assess the independence of certain events?
H.W. 335 – 336 #80, 83, 85 , 87
Do Now: in your own words, given an example or define two events that are independent of each other?

19. Conditional Probability and Independence
When knowledge that one event has happened does not change the likelihood that another event will happen, we say that the two events are independent.
Two events A and B are independent if the occurrence of one event does not change the probability that the other event will happen. In other words, events A and B are independent if
P(A | B) = P(A) and P(B | A) = P(B).
When events A and B are independent, we can simplify the general multiplication rule since P(B| A) = P(B).
Multiplication rule for independent events
If A and B are independent events, then the probability that A and B both occur is
P(A ∩ B) = P(A) • P(B)

20. Ex. – Ebook:
Lefties down under:

21. Practice:
Is there a relationship between gender and relative finger length? To find out, we used the random sampler at the United States CensusAtSchool Web site ( to randomly select 452 U.S. high school students who completed a survey. The two-way table shows the gender of each student and which finger was longer on their left hand (index finger or ring finger).
Problem: Are the events “female” and “has a longer ring finger” independent? Justify your answer.
Female
Male
Total
Index finger 78 45
123
Ring finger 82
152
234
Same length 52 43 95
212
240
452

22. Multiplication Rule for Independent Events
Following the Space Shuttle Challenger disaster, it was determined that the failure of O-ring joints in the shuttle’s booster rockets was to blame. Under cold conditions, it was estimated that the probability that an individual O-ring joint would function properly was Assuming O-ring joints succeed or fail independently, what is the probability all six would function properly?
P( joint 1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK)
By the multiplication rule for independent events, this probability is:
P(joint 1 OK) · P(joint 2 OK) · P (joint 3 OK) • … · P (joint 6 OK)
= (0.977)(0.977)(0.977)(0.977)(0.977)(0.977) = 0.87
There’s an 87% chance that the shuttle would launch safely under similar conditions (and a 13% chance that it wouldn’t).

23. Practice – perfect game.
In baseball, a perfect game is when a pitcher doesn’t allow any hitters to reach base in all nine innings. Historically, pitchers throw a perfect inning—an inning where no hitters reach base—about 40% of the time ( So, to throw a perfect game, a pitcher needs to have nine perfect innings in a row.
Problem: What is the probability that a pitcher throws nine perfect innings in a row, assuming the pitcher’s performance in an inning is independent of his performance in other innings?
Solution:

24. Continued: Solution – continued:
Solution: The probability of having nine perfect innings in a row is
P(inning 1 perfect and inning 2 perfect and and inning 9 perfect)
= P(inning 1 perfect) P(inning 2 perfect) P(inning 9 perfect)
= (0.4)(0.4) (0.4) = (0.4)9 =
On a side note:
With 30 teams playing 162 games per season, this means we would expect to see about (30)(162)( ) = 1.3 perfect games per season. However, in baseball history, perfect games are much rarer, occurring once every 4 or 5 seasons, on average. This discrepancy suggests that our assumption of independence may not have been a good one.

25. At least and none:
E book Rapid HIV testing.

26. Practice:
The First Trimester Screen is a noninvasive test given during the first trimester of pregnancy to determine if there are specific chromosomal abnormalities in the fetus. According to a study published in the New England Journal of Medicine in November 2005 ( approximately 5% of normal pregnancies will receive a positive result.
Problem: If 100 women with normal pregnancies are tested with the First Trimester Screen, what is the probability that at least 1 woman will receive a positive result?
BTW – you may assume that the test results for different women are independent.

27. Solution:
To find the probability of at least one false positive, we can use the complement rule and the probability that none of the women will receive a positive test result:
P(at least one positive) = 1 – P(no positive results)
For women with normal pregnancies, the probability that a single test is negative is 1 – 0.05 = The probability that all 100 women will get negative results is (0.95)(0.95) (0.95) = (0.95)100 = Thus, P(at least one positive) = 1 – = There is over a 99% probability that at least 1 of the 100 women with normal pregnancies will receive a false positive on the First Trimester Screen.

28. Conditional Probabilities and Independence
CALCULATE and INTERPRET conditional probabilities.
USE the general multiplication rule to CALCULATE probabilities.
USE tree diagrams to MODEL a chance process and CALCULATE probabilities involving two or more events.
DETERMINE if two events are independent.
When appropriate, USE the multiplication rule for independent events to COMPUTE probabilities.