MAT 1000 - Wayne State Universitymath.wayne.edu/~brcs/classes/m1000w15/lectures/Lecture9.pdfSuppose...

MAT 1000

Mathematics in Today's World

Last Time

We talked about computing probabilities when all of the

outcomes of a random phenomenon are “equally likely”

We also looked at a way to experimentally determine

probabilities using “simulations”

Today

When all outcomes are equally likely, finding probabilities

only requires counting.

We will learn methods to count even when there are more

outcomes then we could write down in a list.

Counting

When the outcomes of a random phenomenon are equally

likely (all have the same probability), then we have the

following formulas:

probability of any one outcome = 1

total number of outcomes

probability of an event= number of outcomes in the event


(Remember that an event is some collection of outcomes.)

Equally likely outcomes

All that these formulas require is counting.

For example, if I want to know the probability of any of the

outcomes from rolling two dice, it’s enough to know that

there are 36. I don’t need to list them all.

This is especially useful when there are too many

outcomes to list.

Formulas for counting

What is the probability of someone guessing a four-digit

PIN (for example for a bank or debit card)?

If we assume that every four-digit PIN is equally likely, then

we can use the formula



Here the total number of outcomes is the number of four-

digit PINs. How many are there?


Listing them all would be tedious and difficult. Can we still

count them?

Let’s compare this with an easier example.

There are 36 possible outcomes for rolling two dice. We

originally found them by making a list. Can we count them

without the list?

Yes: there are six possible numbers for the first die, and six

for the second, and 6 ⋅ 6 = 36


This is an example of what the book calls the “fundamental

principle of counting.”

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

We have 6 possible outcomes for rolling the first die, and 6

for the second, so there are a total of 6 ⋅ 6 = 36 outcomes.


What about a four-digit PIN?

There are 10 choices for the first digit:

0 1 2 3 4 5 6 7 8 9

Likewise we have 10 choices for the second, 10 for the

third, and 10 for the fourth. So the number of four-digit PINs

is

10 ⋅ 10 ⋅ 10 ⋅ 10 = 10,000


Then the probability that someone can randomly guess

your PIN is

1

total number of outcomes=

1

10,000


If four-digit PINs are randomly assigned, what is the

probability that you get a PIN with four identical digits?

Since each PIN is equally likely, we can use the formula:



We already know that there are 10,000 different four-digit

PINs.

We need to know the number of outcomes in our event.


In this case the number turns out to be small enough to

count without a formula.

Our “event” is getting a PIN all of whose digits are the

same. For example 1111, or 7777. How many such four-

digit PINs are there?

There are ten.

The digits could all be any one of the ten digits 0, 1, … , 9.


Alternately, we can just list all four-digit PINs with all four

digits identical:

So the probability of getting a PIN with four identical digits

is:

number of outcomes in the event

total number of outcomes=

10

10,000=

1

1000

0000 1111 2222 3333 4444

5555 6666 7777 8888 9999


Here’s a different counting problem: how many four-digit

PINs are there if we are not allowed to repeat digits?

In other words a PIN like 1234 is allowed, but not a PIN like

2452 or 7674, where a digit is repeated.

We already know that the number of such PINs must be

less than 10,000, but how many are there?

We can use the fundamental principle of counting.


We are “choosing” four digits. How many choices are there

for the first?

There are 10 choices.

What about the second?

Now, we don’t have 10 choices—we can’t repeat the first

digit.

Suppose the first digit was 7. If we pick 7 for the second

digit, we’ve repeated ourselves. So the second digit can be

any number except 7.

Then how many choice are there?

There are 9 choices—anything except our first digit.


So we have 10 choices for the first digit and 9 for the

second. How many choices are there for the third?

We have already picked two digits, and we can’t reuse

either of those. So we only have 8 choices.

We want a four-digit PIN, so we need to pick one more

number. Since it can’t be the same as the first, second, or

third choice we’ve made, we’ve only got 7 options.


The fundamental principle tells us that there are

10 ⋅ 9 ⋅ 8 ⋅ 7

ways to pick a four-digit PIN with no repeated digits.

Multiplying out, we see there are 5040 such PINs.

One way of looking at what we’ve done:

We have four “spaces”

___ ___ ___ ___

which we want to “fill,” each with a different number.


What’s the probability that if we randomly choose a four-

digit PIN with no repeated digits, then we get the PIN

1234?

Since any PIN is equally likely, the probability of the one

outcome 1234 is



This is 1

5040= 0.002

Counting

What’s the probability that if we randomly choose a four-

digit PIN with no repeated digits, the first digit of our PIN is

1?

What does this mean? We want the first digit to be 1, so

we want a number like 1234 or 1573, but we don’t want

numbers like 3612, where the 1 is not the first digit.

This is an event. So we can use the formula



Counting

How many outcomes are in the event? In other words, how

many PINs are there where no digits are repeated and the

first digit is 1?

We have to count.

Unlike earlier, we don’t have four spaces to fill.

___ ___ ___ ___

because we know what goes in the first space…

Counting

How many outcomes are in the event? In other words, how

many PINs are there where no digits are repeated and the

first digit is 1?

We have to count.

Unlike earlier, we don’t have four spaces to fill.

_1_ ___ ___ ___

because we know what goes in the first space…

… a 1.

Counting

So we only have three spaces to fill.

How many choices do we have for the first space?

Even though there are 10 digits, remember that the first

digit must be a 1, and we aren’t allowed to repeat. So we

only have 9 choices.

How about for the second digit?

There are 8 choices.

Counting

For the third digit we have 7 choices.

So the number of four-digit PINs with no repeated digits

that start with a 1 is:

9 ⋅ 8 ⋅ 7 = 504

Then the probability that we pick a four-digit PIN with no

repeated digits, and we get a PIN that starts with a 1 is:

504

5040=

1

10

Counting

The difference between the number of four-digit PINs

10 ⋅ 10 ⋅ 10 ⋅ 10 = 10,000

and four-digit PINs with no repetitions

10 ⋅ 9 ⋅ 8 ⋅ 7 = 5040

is an important difference.

In both cases we are picking four digits out of the 10

possible, but if we aren’t allowed to repeat, we will have

fewer possible four-digit PINs.

Counting

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.

If we are not allowed to repeat, we have 𝑛 choices for the

first thing, 𝑛 − 1 for the second, 𝑛 − 2 for the third…

and it turns out we will have 𝑛 − 𝑘 + 1 for the last.

As a formula, this is:

𝑛 ⋅ 𝑛 − 1 ⋅ 𝑛 − 2 ⋅ … ⋅ (𝑛 − 𝑘 + 1)

Counting

It’s probably easier not to try and memorize a formula.

Instead, remember it this way: we are picking 𝑘 things, so

we have 𝑘 spaces to fill. There are 𝑛 choices for the first

thing, and then for each subsequent thing, we have one

fewer choice.

An arrangement of 𝑘 things out of 𝑛 possibilities with no

repetitions is called a permutation, which is sometimes

abbreviated:

𝑛𝑃𝑘

Counting

Suppose 10 runners are competing in a race. The winner

gets a gold medal, second place gets a silver medal, and

third place gets a bronze medal. How many different

arrangements of gold, silver, bronze are possible?

Here we are “filling” three “spaces”

______ ______ _______

gold silver bronze

How many possible gold medal winners are there? 10.

Once we know who wins the gold, we’ve only got 9 choices

for silver, and then 8 for the bronze.

Counting

Notice, this is all we care about. We aren’t interested in

fourth place, fifth place, and so on.

So we have

10 ⋅ 9 ⋅ 8 = 720

possible lists of gold, silver, bronze.

By the way, this is exactly

10𝑃3

Counting

What if we do want to know the rest of the places: who

came in fourth, and fifth, and so on?

Now we are filling all 10 spots. We get 10 choices for first

place, 9 for second, 8 for third, 7 for fourth, and so on.

Once we have picked first place through ninth place, we

only have one runner left for last place. So there are

10 ⋅ 9 ⋅ 8 ⋅ … ⋅ 2 ⋅ 1 = 3,628,800

orderings of all 10 runners

Counting

You may know another name for the number

10 ⋅ 9 ⋅ 8 ⋅ … ⋅ 2 ⋅ 1

This is called “10 factorial,” which is usually abbreviated

with an exclamation mark:

10! = 10 ⋅ 9 ⋅ 8 ⋅ … ⋅ 2 ⋅ 1

It turns out that if we pick 𝑛 things from a list of 𝑛 and don’t

repeat, the number of choices is exactly 𝑛!

In other words: 𝑛𝑃𝑛 = 𝑛!

(This is 𝑛𝑃𝑘 when 𝑘 and 𝑛 are the same. When they are

different the formula for 𝑛𝑃𝑘 is more complicated).

Counting

If I was one of the runners, I might like to try and find the

probability of getting a medal.

This is really three “events”

1. I get the gold medal

2. I get the silver medal

3. I get the bronze medal

These events are disjoint (I can’t win more than one

medal), so to find the probability of any one of these three

events occurring, I can add the probabilities of each.

Counting

Let’s count the number of possible arrangements in which I

am the gold medal winner.

__Me___ ______ _______

gold silver bronze

There are 9 runners other than me, and we can put any

one of them in the silver medal spot. That leaves 8 choices

for the bronze medal spot.

So there are 72 arrangements in which I am the gold

medal winner

Counting

Likewise there are 72 arrangements in which I win the

silver medal, and 72 in which I win the bronze medal.

So I might like to conclude that the probability of my

winning a medal is:

72

720+

72

720+

72

720=

216

720=

3

10

But I would be wrong.

Counting

In determining the probabilities of winning a gold, silver, or

bronze medal, I was using the following formula



This formula only works when all outcomes have the same

probability.

In a race, that’s not the case. If I’m racing against Usain

Bolt, he has a much higher probability of winning than I do.

Summary

We have two formulas, both derived from the fundamental

principle of counting.

For both formulas, we are choosing 𝑘 things out of a list of

𝑛

1. If we are allowed to repeat, then we have 𝑛𝑘 ways to

choose our 𝑘 things.

Note that because we can repeat, we can have 𝑘 as large

as we like.

Summary

Example

Computers represent data with 1s and 0s. A single digit (1

or 0) is called a bit. A byte is 8 bits. How many possible

different bytes are there?

We are choosing 8 things from a list of 2, and repetitions

are allowed. So there are

28 = 256

different possible bytes.

Summary

If we are choosing 𝑘 things out of a list of 𝑛, we are not

allowed to repeat, and the order in which we choose

matters, then there are 𝑛𝑃𝑘 ways of choosing.

Note that because we can’t repeat, the number of things

we choose (𝑘) can’t be larger than the number of things to

choose from (𝑛).

MAT 1000 - Wayne State Universitymath.wayne.edu/~brcs/classes/m1000w15/lectures/Lecture9.pdfSuppose...

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