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. 5.3a h.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 StatisticsCalculus Computer Science

(Statistics or Computer Science) and not Calculus StatisticsCalculus Computer Science

Com Sci Statistics and Computer Science and not Calculus StatisticsCalculus Computer Science

Statistics and not (Computer Science or Calculus) StatisticsCalculus 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 coin then 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. HT We follow the branches out to show an outcome.

Example: Picking Two Lefties The 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(1 st student lefty) x P(2 nd student is lefty/1 st student lefty)

Multiplication Principle (for independent events).  A second technique is to use the Multiplication Principle.  If you can do Event A: a ways and 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 is easy.  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 4hds TTTT 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 of heads 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 large to list all of the possible outcomes. If you polled a SRS of 1500 people with a yes or no answer, the number of possible outcomes is: outcomes  Too large to list.

Replacement Selecting objects from a collection of distinct choices such as drawing playing cards, with or with out replacement is important. Draw a card replace, draw another.  Possible outcomes: Draw a card, don’t replace, and draw another.  Possible outcomes: 52 ∙ ∙ 51

Ex: How many 3 digit #’s can we make? Possibilities for 1st, 2nd, and 3rd digit are: 10 ∙ 10 ∙ 10 = 1,000 Possibilities for 1st, 2nd, and 3rd digit w/out replacement are: 10 ∙ 9 ∙ 8 = 720

If 2 events are disjoint (mutually exclusive) then, P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 coin Event 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: RRoll a yellow die and a red die. EEvent A is the yellow die landing on an even number, and event B is the red die landing on an odd number. TThese two events are independent, because the probability of A does not change the probability of B.

Multiplication rule for Independent Events Rule 5: IIf events A and B are independent, then knowing that one occurs does not change the probability that the other occurs, and oToThe 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(R 1 ) x P(R 2 ) = 26/52 ∙ 25/51 =

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: IIndependent Take IQ test twice: N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: yellow For offspring:  GG: green  all other combinations are yellow. P(G M and G F ) = P(G M )P(G F ) (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 diagram because 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 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 of functioning with out failure for 25 years.  Repeaters fail independently of each other.

Denote A i : event that the i-th repeater operates successfully for 25 years.  P(A i ) = Find the probability 2 repeaters both last 25 years.  P(A 1 and A 2 ) = P(A 1 ) ∙ P(A 2 )  = x = 0.998

Is 99.9% reliability good enough? Find the probability 10 repeaters both last 25 years. P(A 1 … A 10 ) = P(A 1 ) P(A 2 ) ….P(A 10 ) = = The last transatlantic cable laid 662 “repeaters”. Probability all 662 work for 25 years: P(A 1 … A 662 ) = = 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.