Lesson 8, Chapter 6.5

Finite Mathematics Lesson 9

Chapters 6.5 - 6.7

Some probabilities are not easy to set up. Warning ! Some of this material is hard. This lesson explores conditional probabilities where we want to find the probability of a event given some other event has already occurred. We also talk independence of events and what that means. Finally, Bayes' formula is introduced but before you panic, I won't make you memorize it. I will show you how to use it. This chapter really makes us think... Sometimes it hurts me !

Course Notes

Section 6.5 Conditional Probability and Independence

Conditional Probability
A coin is tossed 3 times. The sample space is

KEY : H T H = Heads (1st toss) Tails(2nd toss) Heads(3rd toss)

H H H	H H T	H T H	H T T
T T T	T T H	T H T	T H H

The probability of 3 Heads is 1/8 = 0.125
The probability of 2 Heads and 1Tail is 3/8 = 1/4 = 0.375
The probability of 1 Head and 2 Tails is 3/8 = 1/4 = 0.375

What is the probability on tossing exactly 2 Heads if we know the first toss is a Head?

We have a new sample space. The sample space for first toss is a Head is
{H H H, H H T, H T H, H T T }
There are 2 ways to toss exactly 2 Heads given the first toss is a Head.
The probability is 2/4 = 0.5 Note this is different that the probability without the extra condition.

What is the probability on tossing 2 Heads and 1Tail if we know the first toss is a Tail?

We have a new sample space. The sample space for first toss is a Tail is
{T T T, T T H, T H T, T H H}
There is only 1 way to toss 2 Heads and 1Tail given the first toss is a Tail.
The probability is 1/4 = 0.25 Note this is different that the probability without the extra condition.

A New Formula
As you can imagine, when the sample space gets large it would be impossible to list all the outcomes.
We need a new formula.

This is read as "The probability of Event F given Event F"

It is sometimes useful to write as (Product Rule)

Let's look at this problem again.
What is the probability on tossing exactly 2 Heads if we know the first toss is a Head?

Let E = tossing exactly 2 Heads
Let F = the first toss is a Head
= tossing exactly 2 Heads AND the first toss is a Head

Pr(F) = 4/8 = 0.5

Let's look at this problem again.
What is the probability on tossing 2 Heads and 1Tail if we know the first toss is a Tail? { T H H }

Let E = tossing 2 Heads and 1Tail
Let F = the first toss is a Tail
= tossing 2 Heads and 1Tail AND the first toss is a Tail

Pr(F) = 4/8 = 0.5

Example #12 page 291

Let E and F be events with Pr(E) = 0.3, Pr(F) = 0.6, and

Find the probabilities:

	From before we know


	Using the fact that Think about it (page 278, practice problem 2, example 2) So rearranging the formula gives 0.6 - 0.2 = 0.4

Independence

Let E and F be events. We say that E and F are independent provided

What does this mean? If the outcome of an event has no affect on the outcome
of the other event , the events are called independent.

Example Toss 2 fair coins.

Find the probability of tossing a Head on first toss and a Tail on the second toss. (HT)
Pr(HT) = Pr(H) × Pr(T) = 1/2 × 1/2 = 1/4 These events are independent.

Example Roll 2 dice.

Find the probability of a 4 on the first die and 6 on the second die. (4,6)
Pr(4,6) = Pr(4) × Pr(6) = 1/6 × 1/6 = 1/36 These events are independent.

These events are independent because coins or dice have no memory. Every event is a new event.
Counter Example

Suppose we pick 2 balls out of an urn without replacement. Let's say 5 are red and 7 are blue.
Pr(red) = 5/12 and Pr(blue) = 7/12
What is the probability of picking a red ball then a blue ball?
Pr(red,blue) = 5/12 × 7/11 , notice it is not 5/12 × 7/12 because there are less balls in the urn on the second pick.
Therefore, these events are not independent.

A final note: If 2 or more events are independent, you can multiply their probabilities without looking at a new sample space or using the formula for conditional probabilities.

Section 6.6 Tree Diagrams

Nothing really new here other than I good way to organize data. Check this out...

Problem #18, page 300 (page 288 in old edition)
A factory has two machines that produces bolts. Machine I produces 60% of the daily output of bolts. , and 3% of its bolts are defected. Machine II produces 40% of the daily output of bolts. , and 2% of its bolts are defected.

What is the probability that a bolt selected at random will be defected?

I used "Bad" instead of defected. A picture is worth a thousand words (or formulas)
Pr(Defected) = 0.6 × 0.03 + 0.4 ×0.02 = 0.018 + 0.008 = 0.026
Note: and independent

If a bolt is selected at random and found to be defected,
what is the probability that it was produce by machine I?

This is a conditional probability,
Pr( produced by machine I (I) given it is defective (D)) = Pr(I | D)
We know Pr(D) = 0.026 from above. So...

Section 6.7 Bayes' Theorem

Bayes' theorem looks very complicated, but, wait... it is complicated. Let's look at
Problem #18, page 300 again.

b. A factory has two machines that produces bolts. Machine I produces 60% of the daily output of bolts. , and 3% of its bolts are defected. Machine II produces 40% of the daily output of bolts. , and 2% of its bolts are defected.

If a bolt is selected at random and found to be defected,
what is the probability that it was produce by machine I?

Let B₁ = produced by Machine I	Let B₂ = produced by Machine II
Pr(B₁) = 0.6	Pr(B₂) = 0.4
Pr( D \| B₁) = 0.03 probability bolt is defective given it was produce by machine I	Pr( D \| B₂) = 0.02 probability bolt is defective given it was produce by machine II
We want to find the probability that bolt selected at random was produced by Machine I. That is Pr( B₁ \| D). This is not the same as Pr( D \| B₁) where the machine was picked before the bolt was selected. Note :I used Pr( I \| D) in the problem above.
By Bayes' theorem	Notice the pattern!

Recall from above, that this is the same solution we got with the tree. (part b)

Example page 305, problem 2. Look at Table #3

An electronic device has six different types of transistors. For each type of transistors, table 3 gives the proportion of the total numbers of transistors of that type and the failure rate(probability of failing within one year). If a transistor fails, what is the probability that it is of type 1.

We want to find Pr( Type 1 | transistor fails) = Pr(Type 1 | F) where F = fails.

or more simply put...

Pr(Type 1| F) = 0.0698

A final note on Bayes' Theorem

The product of the probability you are looking for goes in the numerator.
The sum of the products of all the probabilities goes in the denominator.

There are additional online tutorials at the books' web site. http://cw.prenhall.com/bookbind/pubbooks/goldstein3/

Please notify me of any errors on this page joe@joemath.com