Conditional Probability and Independence

Target Goals:

I can use a tree diagram to describe chance behavior.I can use the general multiplication rule.I can determine whether two events are independent.

5.3ah.w: p 329: 57 – 60, 63, 77, 79, 87

Warm Up: Practice with Venn Diagrams

On the following four slides you will find Venn Diagrams representing the students at your school.Some students are enrolled in Statistics, some in Calculus, and some in Computer Science.

For the next four slides, indicate what relationships the shaded regions represent. (use complement, intersection, and union)

Calculus or Computer Science

Statistics Calculus

Computer Science

(Statistics or Computer Science) and not Calculus

Statistics Calculus

Computer Science

Com Sci

Statistics and Computer Science and not Calculus

Statistics Calculus

Computer Science

Statistics and not (Computer Science or Calculus)

Statistics Calculus

Computer Science

Example: Flip a Coin and Roll a Die (I can use a tree diagram to describe chance behavior)

What is the sample space of flipping a cointhen rolling a die? One technique is to draw a tree diagram

and count possible outcomes.

Suppose two coins are flipped. The sample space would be:

S = {HH, HT, TH, TT}Where H = heads and T = tails

H

T

H

T

H

T

We can also use a tree diagram to represent a sample space.

HTWe follow the branches out to show an

outcome.

Example: Picking Two LeftiesThe two way table showing the gender and

handedness of the students in Mr. Tabors AP Statistics class is represented below.

Suppose we choose 2 students at random.•Draw a tree diagram that shows the sample space

for this chance process.•Find the probability that both students are

left handed.

P(two lefties) = P(1st student lefty) x P(2nd student is lefty/1st student lefty)

4 30.016

28 27

Multiplication Principle (for independent events).

A second technique is to use the Multiplication Principle.

If you can do Event A: a waysand Event B: b ways

Then you can do, Both: a x b number of ways

Multiplication Principle

There are 2 ways to toss a coin and 6 ways to roll a die so there are,2 x 6 = 12 ways to toss a coin and then roll a die.

Example: Flip Four Coins

Finding the number of outcomes iseasy. 2 x 2 x 2 x 2 = 16

Listing the outcomes is the challenge.List the possible out comes when tossing a coin 4 times and check with a neighbor. Report back:

Possible Outcomes

0 heads 1 head 2 hds 3 hds 4hdsTTTT HTTT HHTT HHHT

HHHH

THTT HTHT HHTH TTHT HTTH HTHH TTTH THHT THHH

THTH

TTHH

Defining is important.

We only want to know the number ofheads in four tosses. Define: Toss coin four times and

count the # of heads.What is the sample space?

S = {0, 1, 2, 3, 4}

Some sample spaces are too largeto list all of the possible outcomes.If you polled a SRS of 1500 peoplewith a yes or no answer, the numberof possible outcomes is:

21500 outcomes Too large to list.

Replacement

Selecting objects from a collection ofdistinct choices such as drawingplaying cards, with or with out replacementis important.

Draw a card replace, draw another. Possible outcomes:

Draw a card, don’t replace, and draw another.

Possible outcomes:

52 ∙ 52

52 ∙ 51

Ex: How many 3 digit #’s can we make?

Possibilities for 1st, 2nd, and 3rd digitare:

10 ∙ 10 ∙ 10 = 1,000

Possibilities for 1st, 2nd, and 3rd digitw/out replacement are:

10 ∙ 9 ∙ 8 = 720

If 2 events are disjoint (mutually exclusive) then,

P(A or B) = P(A U B) = P(A) + P(B)

s/a finding the probability at least one of these events occurs.

Independence

Event A: toss first coinEvent B: toss second coin

Does the first toss affect the 2nd?No, the events are independent.

Independent

Events A and B are independent if knowing that one occurs does not change the probability that the other occurs.

Example:

Roll a yellow die and a red die. Event A is the yellow die landing on an

even number, and event B is the red die landing on an odd number.

These two events are independent, because the probability of A does not change the probability of B.

Multiplication rule for Independent Events

Rule 5: If events A and B are independent, then

knowing that one occurs does not change the probability that the other occurs, and

o The probability of A and B equals the probability of A multiplied by the probability of B.P(A and B) = P(A) ∙ P(B)

Example: The probability than the yellow die lands

on an even number and the red die land on an odd number is:

½ ∙ ½ = ¼ The P (draw 2 red cards) = Is this event independent?

You can still use the multiplication rule, just be careful.

= P(R1) x P(R2)= 26/52 ∙ 25/51= 0.2451

Venn Diagram (displays outcomes)

Venn diagram showing the event {A and B} as the overlapping area common to both A and B.

Independent or Dependent?

Take your blood pressure twice: Independent

Take IQ test twice: Not independent; the first test

gives you information and knowledge.

Example: Mendel’s Peas

Gregor Mendel used garden peas in some of the experiments that revealed that inheritance operates randomly.

Two parents carry two genes and pass on one.

Gene Crossing

G: green, Y: yellowFor offspring: GG: green all other combinations are yellow.

P(GM and GF) = P(GM)P(GF) (0.5)(0.5) = 0.25 ¼ of all seeds produced by crossing will

be green.

Important Note:

P(A and B) = P(A) ∙ P(B) holds for independent events.P(A or B) = P(A) + P(B) holds for disjoint events.

Disjoint events are not independent. (DENI)

If A and B are disjoint, then the fact A occurs tells us that B can’t occur.

If events A and B are independent, then knowing that one occurs does not change the probability that the other occurs.

It doesn’t mean that the other “can’t” occur.

Independence can not be pictured by a Venn diagrambecause it involves the probability of an event,not the outcome which the Venn diagram shows.

Ex: Atlantic Telephone Cable

The first successful transatlantic telegraph cable was laid in 1866. The first telephone cable across the Atlantic did not appear until 1956 – the barrier was designing “repeaters”, amplifiers needed to boost the signal that could operate for years on the sea bottom. The first fiber optic cable was laid in 1988 and had 109 repeaters.

Repeaters

Repeaters in undersea cables must be very reliable. To see why, suppose that each repeater has a probability 0.999 of functioning with out failure for 25 years.

Repeaters fail independently of each other.

Denote Ai: event that the i-th repeater operates successfully for 25 years.

P(Ai) = 0.999

Find the probability 2 repeaters both last 25 years.

P(A1 and A2) = P(A1) ∙ P(A2) = 0.999 x 0.999 = 0.998

Is 99.9% reliability good enough?

Find the probability 10 repeaters both last 25 years.P(A1 … A10) = P(A1) P(A2) ….P(A10)

= 0.99910 = 0.990

The last transatlantic cable laid 662 “repeaters”. Probability all 662 work for 25 years:P(A1 … A662) = 0.999662 = 0.516!

Conclusion:

This cable will fail to reach its’ 25 year design life about ½ the time even if each “repeater” is 99.9% reliable.

Repeaters must be much more than 99.9% reliable.