Download Probability II

PHP 2510
Probability Part 2
Topics to be covered today:
• Multiplication principle
• Counting permutations and combinations, and associated
probability calculations using these
• Binomial and multinomial coefficients
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Last lecture, we defined sample space
Ω = {ω1 , ω2 , ..., ωn },
and event which is a subset of Ω
A ⊆ Ω.
Today, we focus on how to compute the probability of event A for
a finite sample space
Pr(A) =?
Specifically, we will learn how to determine Pr(A) by counting.
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One example
Suppose that a fair coin is thrown twice and the sequence of heads
and tails is recorded. The sample space is
Ω = {hh, ht, th, tt}.
Let A denote the event that at least one head is thrown. Then,
A = {hh, th, ht}.
The fact is that all four sequences have equal probability.
Therefore by counting, the probability of the event A is
Pr(A) = 3/4 = 0.75.
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If the element of Ω all have equal probability and event A can only
happen in mutually exclusive ways, then
Pr(A) =
# of ways A can occur
.
total # of outcomes
• If Ω contains N elements, then each element has probability
1/N .
• If A can only happen in n exclusive ways, then
Pr(A) = n/N.
• Again, the formula holds only if all the outcomes are
equally likely.
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Counting
Now, the question simplifies to how to calculate the total # of
outcomes – e.g. How many different sequences if we throw a
fair coin 100 times? – and the # of ways A can occur.
Multiplication principle. Consider observing a sequence of q
experiments. The first experiment has n1 possible outcomes,
the second one has n2 possible outcomes, and so forth, up to nq
outcomes for the qth experiment. The total number of possible
outcomes is
n1 × n2 × · · · × nq
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Example. Toss a coin 10 times. The number of possible sequences
is 2 × 2 × · · · × 2 = 210 = 1024.
Example. There are 15 teams in the AL and 16 in the NL. Two
teams will play in the world series. The number of possible
world series matches is 15 × 16 = 240.
Example (Driving to work). The number of possible outcomes of
going through 3 intersections is 23 = 8.
Example (Twin). The number of possible gender combinations is
2 × 2 = 4.
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Probability arises from random sampling. Say, sample r objects
from a population of n objects.
This can be envisioned as sampling from an urn containing n
uniquely labeled balls.
There are two ways to do the sampling: With replacement, and
without replacement.
Using the multiplication principle, we have the following
properties of permutation.
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Counting: Permutations
Permutations are ordered samples.
Property 1. The number of ways to order a sample of n objects is
n! = n(n − 1)(n − 2) · · · 1.
In other words, there are n! ways to sample n balls from an urn of
n labeled balls without replacement.
Example. How many ways can 3 individuals be lined up?
Example. How many ways can 10 individuals be lined up?
10! = 3, 628, 800.
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Property 2. Consider a set of n objects. Sample r of them
without replacement. The number of ways to order them is
n(n − 1)(n − 2) · · · (n − r + 1)
Example. From 10 children, five are to be chosen and lined up.
How many different lines are possible?
10 × 9 × 8 × 7 × 6 = 30, 240.
Example. The Powerball game has 55 numbered balls in a drum.
Five are selected without replacement. How many unique
orderings are there? Ans = 417,451,320
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Property 3. If the r objects are selected with replacement,
then the number of ways to order them is
n × n × ... × n = nr .
Example. How many unique orderings of the Powerball numbers
are there if they are selected with replacement? Ans =
503,284,375
Example. In some states, license plates have 6 characters: 3
letters followed by 3 numbers. How many distinct such plats
are possible.
Example. What is the probability that the license plate for a new
car contains no duplicate letters or numbers?
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Birthday probabilities
Our class has 25 people. What is the probability that at least two
individuals have the same birthday? (Ignore leap years).
Define A as the event that at least two people have same birthday
Sometimes easier to work with complement: Ac is the event that
no two people have same birthday, or that there are 25 unique
birthdays
Think of the sample as being drawn from a set of size 365. Each
element has an equal probability.
In this sample of size 25, the probability of Ac is number of ways
to have 25 unique birthdays divided by total number of
possible birthdays. So n = 365 and r = 25.
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To count Ac , how many ways to sample 25 birthdays without
replacement?
365 × 364 × · · · × (365 − 25 + 1)
To count total number of possible birthdays, how many ways to
sample 25 birthdays with replacement?
36525
Probability of 25 unique birthdays (no matches):
365 × 364 × · · · × (365 − 25 + 1)
≈ 0.43
25
365
Probability of at least one match is 1 − P (Ac )
1 − 0.43 = 0.57
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Counting: Combinations
Combinations are unordered samples, usually thought of as being
sampled without replacement.
In particular, we ask the following question: If r objects are taken
from a set of n objects without replacement and disregarding
order, how many different samples are possible?
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Property. The number of possible combinations of r objects
drawn from a set of n objects is
n(n − 1)(n − 2) · · · (n − r + 1)
n!
=
r!
r!(n − r)!
• The numerator is the number of ways to select r objects in
order (number of permutations).
• The denominator is the number of ways to order them once
they are selected.
Notation:
µ ¶
n!
n
=
r
r!(n − r)!
This is called the binomial coefficient
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Example 1: Taste testing.
A statistics professor claims that he can distinguish between three
varieties of cabernet. Given three unlabeled glasses, what is the
probability he can classify them correctly just by guessing? What
is the probability if there are four different varieties?
Ans: 1/6, 1/24
Suppose he classifies them correctly. How do you judge his claim?
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Example 2: Powerball.
Powerball is played in the following way.
• First, five white balls are selected without replacement from a
drum with 55 balls labeled 1 through 55.
• Next, one red ‘powerball’ is selected from a drum with 42 balls
labeled 1 through 42
• If you match the five white balls in any order, plus the red one,
you win.
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Here are the questions:
1. How many combinations are possible for the five white balls?
Ans = 3,478,761
2. The prize for selecting all five white balls correctly is 200K. If I
select the numbers 1-2-3-4-5 for the white balls, what is my
probability of winning the 200K? Is this different than the
probability for any other number?
3. How many combinations are possible for the five white balls
plus the powerball?
Ans = 146,107,962
4. How many people need to play before we can expect a winner?
powerball.com
www.durangobill.com/PowerballOdds.html
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