Transcript MAT 1000 - Wayne State University

MAT 1000
Mathematics in Today's World
Last Time
We learned some formulas for quickly counting large
numbers of things.
The first, called “the fundamental principle of counting” is:
If there are 𝑎 ways of choosing one thing, 𝑏 ways of
choosing the next, and so on up to 𝑧 ways of choosing the
last, then altogether the number of choices is
𝑎 ⋅ 𝑏 ⋅ …⋅ 𝑧
Last Time
Another formula we saw is a special case of this
fundamental principle.
If we are choosing 𝑘 things from of a list 𝑛 possibilities, and
we are allowed to repeat, then there are
𝑛 ⋅ 𝑛 ⋅ … ⋅ 𝑛 = 𝑛𝑘
possible choices.
Last Time
Another formula is used for counting “permutations”
Choosing 𝑘 things out of 𝑛 possibilities, when we can’t
repeat (choose the same thing twice), and it matters what
order we pick in.
This is sometimes abbreviated 𝑛𝑃𝑘
A formula for finding 𝑛𝑃𝑘 is:
𝑛 ⋅ 𝑛 − 1 ⋅ 𝑛 − 2 ⋅ … ⋅ (𝑛 − 𝑘 + 1)
Today
We will introduce another counting formula.
We will discuss the idea of “expected value,” also called
“probability mean.”
Expected value is the reason casinos and insurance
companies can turn a profit. They rely on the “law of large
numbers.”
Counting
One of the lottery games in Michigan is called “Fantasy 5.”
Players select 5 numbers from 1 to 39. Then 5 numbers
are drawn at random—if a player’s numbers match the five
drawn, they win the jackpot.
Can we find the probability of winning?
Any list of five numbers is equally likely to be the winning
numbers, so
1
probability of any one outcome =
total number of outcomes
How many lists of five numbers are there?
Counting
We are trying to count ways of filling five “spaces”
____
____
____
____
____
In each space, we have 39 choices, and we don’t repeat.
But we can’t use the formula for permutations.
In a four-digit PIN or a list of gold, silver, bronze medal
winners, the order of the list matters. The PIN 1234 is not
the same as the PIN 2413.
This isn’t true of lottery numbers.
Counting
If you pick the five numbers 10 15 7 38 26
and the winning numbers are 7 10 15 26 38, then you
win the jackpot.
So if we use the formula for permutations:
39𝑃5
= 39 ⋅ 38 ⋅ 37 ⋅ 36 ⋅ 35
we are overcounting, because we are counting 10 15 7
38 26 and 7 10 15 26 38 separately, when they are
actually the same.
Counting
If the winning numbers are 7 10 15 26 38, then it turns
out there are 5! different orderings of these five numbers
(you can use the formula for permutations to see this:
5𝑃5 = 5! ).
In fact, for any list of five different numbers, we could
reorder those numbers in 5! ways.
So we have to take the number of permutations 39𝑃5 , and
divide it by 5!
Counting
The number of possible lists of 5 different numbers from 1
to 39, where the order of the numbers doesn’t matter is:
39 ⋅ 38 ⋅ 37 ⋅ 36 ⋅ 35
39 ⋅ 38 ⋅ 37 ⋅ 36 ⋅ 35
=
5!
5⋅4⋅3⋅2⋅1
This is equal to 575,757
There is a formula in general for this kind of calculation.
Counting
As in the earlier examples, we are choosing 𝑘 things from
of a list 𝑛 possibilities
We are not allowed to repeat (pick the same thing twice),
but unlike for permutations, the order in which we pick
things does not matter.
To count the number of ways to do this, we divide the
number of permutations 𝑛𝑃𝑘 by 𝑘!
𝑛 ⋅ 𝑛 − 1 ⋅ 𝑛 − 2 ⋅ … ⋅ (𝑛 − 𝑘 + 1)
𝑘 ⋅ 𝑘 − 1 ⋅ …⋅ 2 ⋅ 1
Counting
An arrangement of 𝑘 things out of 𝑛 possibilities with no
repetitions where the order does not matter is called a
combination.
This is usually abbreviated:
𝑛𝐶𝑘
Sometimes people read this abbreviation as “n choose k.”
It is also written:
𝑛
𝑘
Counting
It may help to do another example.
Let’s find 17𝐶6 .
This is the number of ways to pick 6 things out of 17
without repeating, and ignoring order.
We make a fraction. In the numerator, we have 17𝑃6 , in
other words, start from 17, and multiply six numbers
decreasing each by 1
Counting
So the numerator is
17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13 ⋅ 12
The denominator of the fraction is 6!. In other words, start
from 6 and multiply six numbers decreasing each by 1:
6⋅5⋅4⋅3⋅2⋅1
So
17 ⋅ 16 ⋅ 15 ⋅ 14 ⋅ 13 ⋅ 12
= 12,376
17𝐶6 =
6⋅5⋅4⋅3⋅2⋅1
Counting
Here’s another formula for combinations:
𝑛𝐶𝑘
=
𝑛𝑃𝑘
𝑘!
Counting
Example
There are 183 students in this MAT 1000 course. If I
choose a random sample of 5 students, how many different
groups of 5 can I get?
No repetitions, order doesn’t matter.
183 ⋅ 182 ⋅ 181 ⋅ 180 ⋅ 179
= 1,618,621,641
183𝐶5 =
5⋅4⋅3⋅2⋅1
Counting
We’ve now got 3 formulas for counting. How do we know
which one applies?
Suppose you are choosing 𝑘 things from a list of 𝑛. To
decide on a formula, ask the following questions:
1. Am I allowed to repeat (pick the same thing more than
once?). If “yes” then use the formula: 𝑛𝑘
2. If “no” then ask if the order in which you choose
matters.
If “yes” then use𝑛𝑃𝑘
If “no” then use 𝑛𝐶𝑘
Counting
In fact there is a fourth counting formula, which is used for
counting when you are allowed to repeat but the order
doesn’t matter (𝑛𝑘 is wrong for this situation).
But we don’t discuss this formula in this class. If you are
interested in seeing it, it is in the textbook.
Counting
Returning to the Fantasy 5 lottery, we have seen that there
are 39𝐶5 = 575,757 possible lists of 5 numbers from 1 to 39.
Since these are all equally likely, the probability that you
win the Fantasy 5 lottery is:
1
1
=
= 0.0000017
total number of outcomes 575,757
Counting
It’s a little hard to understand probabilities this small. So
here’s a way to make this more concrete.
575,757 is roughly how many seconds there are in one
week.
Imagine that I choose one specific second over the next
seven days. Winning the Fantasy 5 lottery is like picking
that exact second.
Expected value
Of course if 500,000 people play, then we do expect
someone to win. The lottery is daily, and the jackpot is at
least $100,000 (it varies).
But even though someone probably wins every day, you
don’t need to worry about the state of Michigan.
Just like casinos and insurance companies, they know how
to price the tickets so that they will always make money.
How do they do that?
Using expected value, which your textbook calls the
probability mean.
Expected value
Suppose the outcomes of a random phenomenon are
numbers. We’ll denote them by 𝑥1 , 𝑥2 , … , 𝑥𝑛 .
Each of these outcomes has a certain probability (they
don’t have to be equally likely). Denote the probability of
𝑥𝑖 by 𝑝𝑖 . So the probability of 𝑥1 is 𝑝1 , the probability of 𝑥2 is
𝑝2 and so on.
The expected value or probability mean is given by
𝑥1 ⋅ 𝑝1 + 𝑥2 ⋅ 𝑝2 + ⋯ + 𝑥𝑛 ⋅ 𝑝𝑛
This is sometimes denoted by 𝜇 (the Greek letter “mu”).
Expected value
What is the expected value of the Fantasy 5 lottery? For
simplicity, assume there are two outcomes: you win a
$100,000 jackpot, or you win nothing.
The probability of winning is
1
575,757
= 0.0000017
What’s the probability of not winning? By the rules of
probability it is
1
1−
575,757
Which is 1 − 0.0000017 = 0.9999983
Expected value
So the expected value is
100,000 ⋅ 0.0000017 + 0 ⋅ 0.9999983
This is 0.17. Since the outcomes are dollars, the unit is the
same. So the expected value is $0.17, or 17 cents.
Knowing expected values is important (for casinos and
insurance companies) because of a fact called the law of
large numbers.
The law of large numbers
According to the law of large numbers, if we repeat many
trials of a random phenomenon and take the average 𝑥 of
the outcomes, the more trials we repeat the closer this
average 𝑥 will get to the expected value 𝜇.
So if you played the Fantasy 5 lotto many times (meaning
a few hundred thousand at least) and take your average
winnings:
total amount won
number of times played
This average will get closer and closer to $0.17
The law of large numbers
This is fairly useless information to you. If you had enough
money to play the lottery hundreds of thousands of times,
why would you?
But this is very important to the state of Michigan. Over the
course of a year, millions of people will play the Fantasy 5.
According to the law of large numbers, Michigan can
expect to pay out on average of $0.17 per player.
So as long as a lotto ticket costs more than $0.17, they will
make a profit.
The law of large numbers
The law of large numbers is also important for casinos.
Again, not for you as a player. You can’t play enough times
for it to matter.
But there will be millions of gambles made at the casino,
and with that many gambles, the average payout will be
very close to the expected value.
So they can rely on the law of large numbers when they set
prices, and they know they will make a profit.
Expected value
Here’s another example of using expected values.
At a carnival, you pay $2.00 to play a coin-flipping game
with three fair coins. On each coin, one side has the
number 0, and the other side has the number 1. You flip
three coins, and you win $1.00 for each 1 that appears on
top.
How do your expected earnings compare to the cost of the
game?
Expected value
First, let’s look at the probability model for tossing these
three coins.
We can think of outcomes as lists of three numbers, which
are either 1 or 0. In other words, we are choosing three
things from two candidates.
Notice we can repeat (all of the coins can come up 1 or 0).
Expected value
How many outcomes are there?
We can use the counting formula 𝑛𝑘 .
Here 𝑛, the number of things we are choosing from, is 2,
and 𝑘, the number of choices we make, is 3.
So there are
possible outcomes.
23 = 8
Expected value
It’s not hard to list them all:
000
100
010
001
110
101
011
111
These are all equally likely. Why?
The coins are fair, meaning the probabilities of getting a 0
1
or a 1 on a single coin toss are equal, both are
2
Also, each of the three coin tosses are independent. So the
1 1 1
1
probability of any of the eight outcomes is ⋅ ⋅ =
2 2 2
8
Expected value
This is the probability model for tossing three coins. Based
on this we can write down a probability model for the
game.
The outcome of the game is amount of money we win,
which equals the number of coins which show a 1.
So the sample space is
{0,1, 2, 3}
Expected value
We win 0 dollars when all three coins show 0. The
1
probability of this is
8
We win 1 dollar when only one of the three coins shows a
1. This can happen 3 ways out of 8 equally likely
3
outcomes, so the probability is
8
3
8
Similarly, the probability of winning 2 dollars is , and the
probability of winning 3 dollars is
1
8
Expected value
This is our probability model
Outcome
Probability
Win $0
1/8
Win $1
3/8
Win $2
3/8
Win $3
1/8
What’s the expected value?
1
3
3
1 12
0⋅ +1⋅ +2⋅ +3⋅ =
= $1.50
8
8
8
8
8
Expected value
So, on average, the carnival pays out $1.50 per game.
However, the cost of playing is $2.00.
So, on average, the carnival gets $0.50 profit per game.
If 1000 people play the game, the carnival will take in
approximately
1000 ⋅ $0.50 = $500