Transcript document

Please turn off cell phones, pagers, etc.
The lecture will begin shortly.
Announcements
• Exam 3 will be held next Wednesday, April 5.
• It will cover material from Lectures 20-27,
or chapters 12, 13 and 16.
• Each student may bring one sheet of notes
(8.5 × 11 inches, both sides) to use during the
exam
• Monday’s class (Lecture 28) will be a review and
cover example questions.
Lecture 27
This lecture will cover additional topics related to
Chapter 16.
1. Rules of probability (Section 16.4-16.5)
2. Expected value (Section 16.6)
1. Rules of probability
Last time, we covered four rules of probability
Rule 0:
0 ≤ P(A) ≤ 1
Rule 1:
P(not A) = 1 – P(A)
Rule 2:
P(A or B) = P(A) + P(B)
if the events A and B are mutually exclusive
Rule 3:
P(A and B) = P(A) × P(B)
if the events A and B are independent
Mutually exclusive
Two events are mutually exclusive if they cannot both
happen
For example, the events
A: it will rain tomorrow
B: it will snow tomorrow
are not mutually exclusive, because it is possible to have
rain and snow on the same day.
On the other hand, the events
A: a person has no siblings
B: a person has three or more siblings
are mutually exclusive, because both cannot be true.
Independent events
Two events are independent if knowing whether one
event has occurred does not change the probability that
the other will occur.
For example, suppose a family has two children (not twins).
The events
A: first child is a girl
B: second child is a boy
are independent, because the sex of the first child has no
influence on the sex of the second child.
The events
A: first child is a girl
B: the family has at least one girl
are not independent, because knowing A has occurred
increases the probability of B.
Comments
• If two events are mutually exclusive, they cannot be
independent.
• If two events are independent, they cannot be
mutually exclusive.
• Don’t fall prey to the gambler’s fallacy.
The gambler’s fallacy arises when a person erroneously
believes that independent events influence each other.
For example, the person may believe that he’s “due” to
win the lottery soon, because he hasn’t won in a long
time.
Repeated addition
The rule
P(A or B) = P(A) + P(B)
can be extended to three or more events, provided that
they are all mutually exclusive (i.e., that no two of them
can occur at the same time).
P(A or B or C or…) = P(A) + P(B) + P(C) + …
For example, if you roll a fair die, the probability of getting
an even number is
P(even number) = P(2 or 4 or 6)
= P(2) + P(4) + P(6)
= 1/6 + 1/6 + 1/6
= 1/2
Repeated multiplication
The rule
P(A and B) = P(A) × P(B)
can be extended to three or more events, provided that
they are all mutually independent (i.e., that no event has
any influence on any other)
P(A and B and C and…) = P(A) × P(B) × P(C) × …
For example, if you toss a coin three times, the probability
of getting “HHH” is
P(H and H and H) = P(H) × P(H) × P(H)
= 1/2 × 1/2 × 1/2
= 1/8
When will it happen?
The rule of repeated multiplication allows us to figure out
the probability that an event occurs for the first time on
the kth trial, for any value of k=1,2,3,…
The probability that the event A occurs for the first time
on trial k is
(k-1) times
P(doesn’t occur) × P(doesn’t occur) × … × P(doesn’t occur)
× P(does occur)
once
Example
If you roll a pair of dice once, the probability of getting a
7 is 6/36 = 1/6.
Suppose you decide to keep rolling until you get a 7.
The probability that you will only have to roll once is
1/6 = 0.167.
The probability that you will have to roll twice is
5/6 × 1/6 = 5/36 = 0.139.
The probability that you will have to roll three times is
5/6 × 5/6 × 1/6 = 25/216 = 0.116.
…and so on.
Probability of happening by a certain time
The rule of repeated multiplication also allows us to figure
out the probability that an event occurs sometime within
the first k trials.
The probability that the event occurs sometime within the
first k trials is
k times
1 – [ P(doesn’t) × P(doesn’t) × … × P(doesn’t) ]
Example
Suppose that, in any year, the earth will be hit by an
asteroid with probability 1/100,000 = .00001.
The probability that it will be hit in the next century is
100
1 – (.99999)
≈ .0001
The probability that it will be hit in the next millenium is
1000
1 – (.99999)
≈ .001
The probability that it will be hit in the next 10,000 years is
10000
1 – (.99999)
≈ .095
2. Expected value
Often the outcome of an experiment is a number.
If the outcome of an experiment is a number, that number
is called a random variable.
Examples of random variables:
• How much you win in a lottery
• The blood pressure of a subject chosen at random
from a population
• The exam score of a student chosen at random
from a class
Probability distribution
The probability distribution for a random variable is
• a list of all possible outcomes, and
• the probabilities of those outcomes
(must add up to one)
For example, suppose that you have a ticket for a
lottery in which you win $1000 with probability 1/5,000
and $1 with probability 1/100. The distribution is:
outcome
1
1,000
0
probability
1/100 = 0.01
1/5000 = 0.0002
1 – .01 – .0002 = 0.9898
Expected Value
To find the expectation or expected value of a random
variable,
• multiply each outcome by its probability, and
• add them up
Example
outcome
1
1,000
0
Prob.
0.01
0.0002
0.9898
outcome × prob
.01
0.2
0
Expected value = .01 + 0.2 + 0 = .21 = $0.21
Interpretation of Expected Value
The expected value is the average value of the outcome
over the long run, if the experiment were repeated many
times.
(For this reason, it is sometimes called the mean.
But don’t confuse it with a sample mean, which we
learned about in Lecture 11.)
In the lottery example, it would be the amount that you
would win on average if you played the lottery over and
over.
It is not the amount that you win if you play just once. (If
you play just once, it’s impossible to win $.21.
Another example
The experiment is to roll a single die. What is the
expected value?
outcome
prob.
1
2
3
4
5
6
1/6
1/6
1/6
1/6
1/6
1/6
outcome × prob.
1/6
2/6
3/6
4/6
5/6
6/6
Expected value = 1/6 + 2/6 + 3/6
+ 4/6 + 5/6 + 6/6
= 21/6
= 3.5