Transcript PowerPoint

MAT 1000
Mathematics in Today's World
Last Time
We discussed the four rules that govern probabilities:
1. Probabilities are numbers between 0 and 1
2. The probability an event does not occur is 1 minus the
probability that it does
3. The probability of one or the other of two disjoint events
occurring is the sum of their probabilities
4. All possible outcomes together have a probability of 1
Last Time
We also discussed “independent” events.
If two events are independent, the probability that both
happen is the product of the probabilities of each.
We also saw how to analyze simple probabilities using sample
spaces, and probability models.
Probability models can be represented visually by probability
histograms
Today
We will talk about random phenomena with “equally likely”
outcomes.
Meaning: all outcomes have the same probability.
When outcomes are all equally likely, to find probabilities we
just need to count.
Today
We will also talk about “simulation.”
Probability is a long-term frequency, so we can experimentally
approximate probabilities.
Equally likely outcomes
A probability model is a list of all possible outcomes (the
sample space) and the probability of each.
Models where all outcomes have the same probability are
especially easy to work with.
When all outcomes in a probability model have the same
probability, we say the outcomes are equally likely.
Equally likely outcomes
Some of the simple examples we have considered have
equally likely outcomes.
Flipping a coin:
Rolling a die:
Equally likely outcomes
But, not every probability model has equally likely
outcomes. We’ve seen a few examples of this.
The gender of a newborn baby:
The color of a randomly drawn M&M
Equally likely outcomes
If we know that a probability model has equally likely
outcomes, finding the probability of each individual
outcome is quite easy.
When all outcomes are equally likely:
1
probability of any one outcome =
total number of outcomes
Equally likely outcomes
Last time we found the sample space for rolling two dice:
I also said that the probability of any one outcome is:
1
36
The reason is that there are 36 possible outcomes, and
they are all equally likely.
Equally likely outcomes
“Outcomes” and “events” are somewhat different things
An event is a collection of outcomes.
When rolling two dice, “the sum of the numbers on the dice
is 10” is an event (not an outcome).
Which outcomes are in this event?
Equally likely outcomes
What is the probability of this event? I claimed last time that
it is
3
1
or
36
18
Why is this true?
When all outcomes are equally likely:
number of outcomes in the event
probability of an event=
total number of outcomes
Equally likely outcomes
Be careful. These two formulas are not always true:
1
probability of any one outcome =
total number of outcomes
number of outcomes in the event
probability of an event=
total number of outcomes
They are only true when all of the outcomes are equally
likely.
Probability myths
Suppose we flip a coin six times in a row. For each flip,
we’ll write down H if we get heads and T if we get tails.
Which of these sequences is less likely to happen?
HHHHHH
THHTHT
Almost everyone would answer that the first sequence is
less likely. This is false.
Probability myths
Each flip of a coin is independent.
Getting heads or tails on one flip doesn’t change the
probabilities of getting heads or tails on the next.
What’s the probability of getting heads on two flips?
1 1 1
probability of H H = ⋅ =
2 2 4
Probability myths
What’s the probability of getting heads on three flips?
1 1 1 1
probability of H H H = ⋅ ⋅ =
2 2 2 8
Four?
1 1 1 1
1
probability of H H H H= ⋅ ⋅ ⋅ =
2 2 2 2 16
Probability myths
By now you see the pattern, and you can guess that the
1
probability of the sequence H H H H H H should be
2
multiplied by itself 6 times.
1 1 1 1 1 1
1
⋅ ⋅ ⋅ ⋅ ⋅ =
2 2 2 2 2 2 64
What is the probability of the sequence T H H T H T ?
Exactly the same.
Probability myths
The probability that we toss a coin and get tails is
1
2
The next toss is independent of this, so the probability of
1
getting heads is
2
Becausae these two tosses are independent, we have
1 1 1
probability of T H = ⋅ =
2 2 4
The next toss is also independent, so
1 1 1 1
probability of T H H = ⋅ ⋅ =
2 2 2 8
Probability myths
Again you should see the pattern. Each of the six tosses is
independent, so the probability of the sequence
THHTHT
1
2
is multiplied by itself 6 times:
1 1 1 1 1 1
1
⋅ ⋅ ⋅ ⋅ ⋅ =
2 2 2 2 2 2 64
The probability of tossing a coin six times and getting
THHTHT
is the same as the probability of getting
HHHHHH
Probability myths
Our intuition is wrong in believing that
THHTHT
is more likely than
HHHHHH
But, there is a kernel of truth in this false intuition.
It is true that tossing a coin six times and getting heads
three times is more likely than tossing a coin six times and
getting heads all six times.
Probability myths
Compare the wording:
“getting heads three times in six tosses”
“tossing a coin six times and getting T H H T H T”
These are different statements. Why?
There are several ways to toss a coin six times and get
heads three times. The sequence T H H T H T is just one
way. Here are two more:
TTHHTH
HHTTTH
Probability myths
So what is the probability of tossing a coin six times and
getting heads three times?
Turns out it’s
20
5
=
64 16
Probability myths
Any sequence of six outcomes for our coin tosses is
equally likely (they all have probability 1/64).
Remember that
number of outcomes in the event
probability of an event=
total number of outcomes
So finding the probability of getting heads on three tosses
requires counting how many ways there are to do this.
Probability myths
Turns out there are 20 (we’ll see a nice formula for
counting this next time) ways to get three heads on six
tosses.
Also there are 64 possible outcomes (another formula we’ll
learn next time).
So, as I said, the probability is
number of outcomes in the event 20
=
total number of outcomes
64
Simulation
This type of question quickly get difficult to analyze.
Suppose we want to know the probability of tossing a coin
ten times and getting a “run” of at least three heads in a
row.
This can be calculated mathematically, but it’s fairly subtle.
Fortunately, there is an alternative approach.
We can estimate probabilities using “simulation.”
Simulation
Probabilities are long-run frequencies.
To estimate the probability of getting a run of three or more
heads in ten coin tosses, we could repeatedly toss a coin
ten times.
Keeping track of how often we get a run of three or more
on ten tosses, we could look at the frequency:
number of times we get a run of three or more heads
number of times we toss the coin ten times
The more times we do this, our approximation will get
closer and closer to the actual probability.
Simulation
But tossing a coin thousands of times would take a long
time.
We can use what’s called “simulation” to speed up the
process (and this can be done on a computer).
The key idea of simulation: use random digits to imitatate
chance behavior.
Simulation
Setting up a simulation:
Step 1. Give a probability model for the underlying random
phenomenon.
Our model is that we have two outcomes: “heads” and
“tails.” They are independent, and equally likely (probability
1
of either is )
2
Simulation
Setting up a simulation:
Step 2. Assign digits to represent outcomes.
If we use a table of random digits, we will get digits from 0
to 9. Each digit is equally likely (probability 0.1), and
successive digits are independent.
Simulation
In our simulation each digit will represent a coin toss.
We can let odd digits represent the outcome “heads” and
let even digits represent the outcome “tails.”
Note this works out because there are 5 odd digits and 5
even digits (including 0), so the probability of choosing an
odd number is exactly the probability of getting heads (0.5),
and the probability of choosing an even number is exactly
the same as the probability of getting tails (0.5 again).
Simulation
Setting up a simulation:
Step 3. Simulate many repetitions.
We want to simulate what happens when toss a coin ten
times. Each digit is one toss. So we look at ten digits from a
table of random digits to simulate ten tosses.
Let’s start from the first line…
Chapter 2
32
Simulation
The first 10 digits are
1922395034
Odd means “heads” and even means “tails,” so these 10
digits simulate tossing a coin ten times and getting:
HHTTHHHTHT
This time we did get 3 heads in a row.
Let’s do two more repetitions.
Simulation
The next 10 digits on the table are
0575628713
This corresponds to the sequence:
THHHTTTHHH
Another sequence with three heads in a row (twice).
The next 10 digits are
9640912531
HTTTHHTHHH
Yet again we have three heads in a row.
Simulation
At this point our simulation estimates the probability of
tossing a coin ten times and getting three or more heads to
be
3
=1
3
Not a very good estimate.
We haven’t done enough simulations. Of course a
computer can quickly do thousands, which is enough to
give a good estimate.
Simulation
If we do 17 more simulations, for a total of 20, heads
occurs at least three times in a row in 11 simulations.
This gives an estimated probability of
11
= 0.52
20
For comparison (using some difficult math) the actual
probability is
0.5078125
So even 20 simulations gives a decent estimate.