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Statistics 60: Fall 2011
Section 4 (Oct 21)
TA: Dennis Sun
Statistics 60: Section 4
1
The Counting Principle
Sample space: the set of all possible outcomes
of a random phenomenon
Event:
any particular set of outcomes.
Shown at right is the sample space for
rolling a die. Shaded is the event that an even
number is rolled.
We assign each outcome a probability between 0 and 1. The probabilities must add to 1. To
find the probability of an event, we simply add the probabilities of the outcomes in that event.
The calculations are particularly simple if, as above, we choose a sample space in which every
outcome has the same probability (in the case above, 1/6). Then, to calculate the probability of
any event, all we need to do is count the number of outcomes in that event:
When working with an equal probability sample space S, the probability of any event A is:
Pr(A) =
# outcomes in A
# outcomes in S
Example 1 (Get Out of Jail). In the board game Monopoly, if you land in jail you must roll a
double (i.e. the two dice show the same number) to get out. What are your chances of getting out
on a given turn?
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Basic Probability Facts
The conditional probability of B given A is written Pr(B|A), and can be calculated by:
Pr(B|A) =
Pr(A and B)
Pr(A)
Two events A and B are independent if:
Pr(B|A) = Pr(B)
Independence captures the notion that A gives no information about B (and vice versa).
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Statistics 60: Fall 2011
Section 4 (Oct 21)
TA: Dennis Sun
Three Handy Rules
Addition Rule: Pr(A or B) = Pr(A) + Pr(B) − Pr(A and B).
We say A and B are mutually exclusive if Pr(A and B) = 0. In that case:
Pr(A or B) = Pr(A) + Pr(B)
Multiplication Rule: Pr(A and B) = Pr(A) Pr(B|A) (same as the conditional probability rule!)
If A and B are independent, then Pr(A and B) = Pr(A) · Pr(B).
Complement Rule: Pr(Ac ) = 1 − Pr(A).
Example 2 (Get Out of Jail Free). In Monopoly, if you are in jail and fail to roll a double in three
turns, then you must pay a fine to be released. About what percentage of the time will you get out
of jail without having to pay a fine?
Example 3 (The Chevalier’s Puzzle). In the 17th century, a French gambler, the Chevalier de
Méré, found it profitable to bet on getting at least one in 4 throws of a die. When predictably
in 24
he ran out of takers for this bet, he devised a new scheme: to bet on getting at least one
throws of two dice. He reasoned that the two games should be the same since:
Pr(at least one
Pr(at least one
1 1 1 1
+ + + =4×
6 6 6 6
1
1
in 24 rolls) =
+ ··· +
= 24 ×
36
36
in 4 rolls) =
1
2
=
6
3
1
2
=
36
3
To his alarm, he began losing money, so he asked his friend, Blaise Pascal (of Triangle fame) to
help him out. Pascal solved the problem, establishing the foundations of probability. Can you?
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Statistics 60: Fall 2011
Section 4 (Oct 21)
TA: Dennis Sun
Example 4 (Blackjack). In the casino game Blackjack, every card is assigned a point value:
ˆ Cards numbered 2-10 are worth as many points.
ˆ Face cards (Jack, Queen, King) are worth 10 points.
ˆ Ace is worth 1 or 11 points (depending on which is better for your cause).
Each player is dealt an initial two card hand, and the objective is to get the sum of the cards as
close to 21 without going over. The hand is called a “blackjack” if the two cards sum to exactly 21.
(a) Suppose you are dealt from a single deck. What is your probability of getting a blackjack?
(b) Alas, you are only dealt two 10’s (which is still pretty good). Now you are left to hope that the
dealer doesn’t get a blackjack. She deals herself two cards (one face up, one face down). The
face up card is an A♠. What is the probability that she has a blackjack, given this information?
Example 5 (Birthday Problem). What’s the chance that two people in our section will have the
same birthday? Let’s find out!
(a) What is the probability that two people share a birthday?
(b) What is the probability that among three people, none will share a birthday?
(c) Find the probability that at least two people in a room of 80 people share the same birthday.
(d) Long ago on “The Tonight Show”, Johnny Carson decided to test out the birthday problem on
his studio audience of about 100 people. He asked how many of them shared his birthday of
October 23. When none did, he was shocked. What went wrong?
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Statistics 60: Fall 2011
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Section 4 (Oct 21)
TA: Dennis Sun
Bayes’ Rule
Remember that to calculate the probability of B given A, we use the formula:
P (B|A) =
P (A and B)
P (A)
Now what if we know P (A|B) and we want to calculate P (B|A)? For example, we might know
that given that a person has a disease, the probability that he tests positive is 90%. But what the
patient wants to know is: given that he tests positive, what is the probability he has the disease?
Bayes’ Rule allows us to flip conditional probabilities:
P (B|A) =
P (A|B)P (B)
P (A|B)P (B) + P (A|B c )P (B c )
This formula may look complicated, but remember that:
P (A|B)P (B) = P (A and B)
So Bayes’ Rule is nothing more than a dressed-up form of the formula for conditional probability:
P (B|A) =
P (A and B)
P (A and B)
=
c
P (A and B) + P (A and B )
P (A)
Example 6 (Lie Detectors). 10% of the employees in a department store have been skimming
money from the cash register. The manager decides to subject all the employees to a lie detector
test. The lie detector goes off 80% of the time when a person is lying, but it also goes off 25% of
the time when a person is telling the truth. Suppose the lie detector goes off for a given worker.
What is the probability that he is lying?
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