c-n m405/vt s4584 lecture notes - …starnes.limfinity.com/c-n m406 lecture notes 1.pdf · c-n m406...

C-N M406 Lecture Notes (part 1)Based on Wackerly, Schaffer and Mendenhall’s

Mathematical Stats with Applications (2010)B. A. Starnes, 7 – 2021

Lecture We will begin our study in Mathematical Stats with the notion of probability. Our text defines this as the belief in the occurrence of a future event (p. 19). We will say, more precisely, that it is man’s guess at what God will do. There are three major interpretations of probability: subjective, relative frequency and classical. Although we will study the classical interpretation in depth, the other two are quite useful in certain situations. For instance, we know that the law of sin that is at work in mankind because of the constant failure of world systems to maintain peace. We can discuss the rules of probability at a later time, but for now we will visualize the concepts using Venn diagrams.

Ex.

Here we have some examples of sets used in probability. We can think of them as dart boards in which the dart can never go out of the rectangle which we call the universal set or set and in which the dart hits at each point with equal likelihood, and the area of the specified set divided by the total set area is the probability of the set being “selected”. The light blue represents (A ∪B)c. The medium blue represents A\B (read A remove B) and B\A respectively. The darker blue represents (A ∩ B). Note finally that

A ∪ Ac = U and A ∩ Ac = φ. []

Ex. Next, consider two events that may happen in time. That both events would occur (not necessarily simultaneously) would be written A ∩ B. That at least one event occurs would be written A ∪ B. That neither occurs would be written

(A ∪ B)c.That exactly one occurs would be written (A\B) ∪ (B\A), or

(A ∩ B)c ∩ (A ∪ B). []

Lecture Next let’s look at a theoretical setup for probability from Khuri’s book Advanced Calculus for Statistics (2003). From p. 13 we see that a probability space is derived from a topological space called a Borel field of events (or σ-field), B. It is defined on a sample space (set) Ω.

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Lecture It has the properties that Ω ∈ B, E and Ec are both in B, and any countable union of events in B also belongs to B. We can now define a new function P(E) such that

P ∈ [0,1], P(Ω) = 1,

and for a countable union of disjoint Ei’s, P = Σ i P(Ei). Not surprisingly, we denote P(E) as the probability of the event E, and call the triple (Ω, B, P) a probability space. Moreover, if we define (p. 13) a new function X: Ω -> A (⊂ ℝ) as the random variable X, we can now see the true form of this entity by noting that P is the function from A (also known as the support of X) to [0, 1] , whereas, in the Borel case P is directly from Ω to [0, 1].

We can consider Ω as the sample space (p. 28), and as on p. 27, certain elements of The Borel field would be simple events (which ones?). If the number of Ei’s is finite or countable, we say that Ω is a discrete sample space (p. 28). One can see now why Venn diagrams are useful tools for describing sample spaces. With all of this in hand we can state that the three axioms given (p. 29) are actually satisfied in this construct.Axioms of Probability: Let Ω be a universal set containing subset E. Then the function P(E) is defined so that the following axioms hold.

1. P(E) ≥ 0.2. P(Ω) = 1.3. If E1, E2, E3 … form a sequence of disjoint events in Ω, then

P(E1 ∪ E2 ∪ E3 …) = Σ i=1∞P(Ei).

Note that for humansP ∈ (0, 1),

but for God P ∈ 0, 1.

Within the Borel probability space let x ∈ ℝ, and set E = ω ∈ Ω: X(ω) ≤ x. E is a member of B (p. 13). Why? Consider the U distribution in conjunction with Borel properties ii and iii and the connection becomes clear. Now write

P(E) = F(x). This is known as the distribution function of X. Additionally it is known as the cdf or cumulative distribution function, or probability function (the prefix “p” in R) and

F(x) = P(X ≤ x).We'll return to this later.

Ex. Consider a random variable X having the following distribution.

x P(X = x) F(x)

0 0.5 .5

1 .25 .75

2 .25 1

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Ex (cont) More specifically

F(x) = .5xIx[0, 1) + .75xIx[1, 2) + Ix[2, ∞). Note the discontinuities in this function. Their existence determines that this is a discrete and not continuous random variable. Think about this. Again, more on this later. []

Many thanks to Justin Lee for noting that the indication function for Lazarus's (brother of Mary and Martha) lifetime, L, is given thusly using ℝ to represent time (yr).

IL(-10.2, 26.9) + IL(27.0, 44.5]. []

Ex Here is another example of a practical application of the cdf. These are the ACT scores of potential college students listed and ordered. In addition to the “point” probability, the cumulative probability is also listed. Can you find the quartiles of this distribution/rv? What shape does this distribution/rv take? How would you find the standard deviation for this distribution/rv?Ask your teacher about computing probabilities using such a chart. []

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Lecture Here are some important dimensional notes. Ω and A can be easily extended to dimension n, either jointly or separately. Also, note that the famous distributions we discuss are for the most part univariate and pertain to quantitative r.v.s. However, very often, qualitative r.v.s (like color) are simply of a higher dimension and can be represented with a vector r.v. X. For further reading on this, go to the instructory website and see VT S4584 Notes I, or C-N M201 Notes 1 (for a simpler view).

Ecclesiastes 9:1111I returned, and saw under the sun, that the race is not to the swift, nor the battle to the strong, neither yet bread to the wise, nor yet riches to men of understanding, nor yet favour to men of skill; but time and chance happeneth to them all.

Ex. Consider members of a class represent the Ω space, and then let the r.v. X assign numbers in A ⊂ ℝ, thereby illustrating the probability concept.Why do we need X? It's necessary to assign a distribution. []

Ex. Let's track the movement of a vehicle through an intersection. In this case

Ω = Right, Straight, Left.Define X: Ω -> A asX(R) = 1, X(S) = 2, and X(L) = 3 (where A = 1,2,3). P(vehicle turns) = P(X is odd) = .67.P(vehicle continues ahead) = P(X is even) = .33. []

Ex. Next consider two balanced coins being tossed, and the upper faces are observed. Ω = HH, HT, TH, TT, and X: Ω -> A is defined as X(HH) = 1, … , X(TT) = 4. Clearly, P(X = 1) = … = P(X = 4) = .25. What is A?

P(E1 = exactly one head) = P(X = 2 or 3) = .5.P(E2 = at least one head) = P(X < 4) = .75.P(E1 ∩ E2) = P(X = 2 or 3) = .5.P(E1 ∪ E2) = P(X < 4) = .75.P(E1

c ∪ E2) = P(X ∈ A) = 1. []

Lecture We will now turn our attention to some basic mathematical probability concepts, and, consequently, the development of some discrete random variables. We want to take up the method of counting possible events. Theorem 2.1 (p. 39) is the precursor to the Fundamental Counting Rule (FCR) (pp. 39, 40). It tells us that an experiment involving n events, each of which has mi outcomes, i = 1, 2, … n, has a total number of outcomes equal to

m1 x m2 … mn = N.

Ex. 20 people are selected and their birthdays are noted. How many results can be obtained in this experiment? The result is easily obtained with the FCR.

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Ex (cont)N = 36520 = 1.7614x1051.

If each person must have a different birthday, then the total number of outcomes is given by

n = 365*364* … *346 = 1.03676x1051outcome.

Consequently, if we want to know the probability that no two people have the same birthday in a group of 20 that are selected randomly, that is given by

P = n/N = .5886. []

Lecture Definition 2.7 (p. 41) tells us that a permutation is an ordered arrangement of objects. If we take r objects from a group of n objects (n ≥ r), the number of possible arrangements is given by

nPr = n!/(n-r)!

Ex. We can see this illustrated in the previous example where n is actually365 P 20 = 365!/345! = 1.03676x1051outcome

as before. []

Lecture A special case occurs when n = r. Here, the number of ways n objects can be arranged is given by n!/0! = n! Now a combination is the same as a permutation except that the order of the objects is disregarded. A set of n objects taken r at a time (without regard for order) (n ≥ r) is given by nCr = n!/(r!(n – r)!).

It is important to know that these computations can be done quickly in R with the following commands.

choose (n, r);factorial (n);

Ex Here is a proof of the combination formula. With the assumptions above, the number of permutations is …

nPr = n!/(n – r)!Ex (cont) However, we now have a different situation in that given a specific group of r objects, we cannot rearrange them to create different permutations. So before where

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we accounted for r! different arrangements to the total, we now add 1. By dividing the total number of permutations, then, by r! we obtain the number of combinations, which is

nCr = n!/(r!(n – r)!).//. []

Ex. An excellent example of these concepts is seen in the famous “Senate Problem”: How many committees of 5 can be formed in the Senate if each state can be represented only once? There are various approaches to the solution, but one that is straightforward is to consider the number of “one state” committees possible... 50 C 5. Note that there are two senators from each state (each of whom could serve on the committee). So each “one state” committee would have 25 = 32 possible arrangements. The final step uses the FCR. So the total number of committees is

(50 C 5)*32 = 67,800,320 com. []

Lecture The best way to differentiate between permutations and combinations is to think of committees. Combinations are instances where a group of people are chosen to form the committee. Permutations, on the other hand, are situations in which the position during the choosing is important. Which do you think this following example is?

Numbers 132Send thou men, that they may search the land of Canaan, which I give unto the children of Israel: of every tribe of their fathers shall ye send a man, every one a ruler among them. Ex. We'll prove that

Σ i=0n(n C i) = 2n.

Begin by observing that 2n = (1 + 1)n = (n C 0)1n10 + (n C 1)1n-111 + … + (n C n)101n

= Σ i=0n(n C i).

And we might add that a practical application of this is seen in the number of ways that a group of people can be put together in smaller groups. If there are 15 people in the overall group, then there are

215 = 32768 possible groupings of size 15 or less. //. []

Lecture Now, the combination formula is but a simpler version of a more complex formula given in Theorem 2.3 (p. 43). This says that a set of n distinct objects can be broken up into k distinct groups containing n1, … nk indistinguishable objects each, in

N = n!/(n1! n2! … nk!) way.

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Actually we can think of this formula as an extension of the permutation formula as well. We might also think of the groups as being boxes where the objects can be placed. So the number of arrangements of the objects into k groups can also be thought of as the number of ways the boxes can be arranged.

Ex. Another excellent example is to think about the word STATISTICS. How many distinguishable ways can the letters in this word be arranged?

Well, there are 5 groups of letters … S, T, C, A and I. We must think of these letters as the “boxes”, and also think of the positions of the word as being the objects in Theo 2.3. With this venue, we can determine that

N = 10!/(3! 3! 2! 1! 1!) = 50,400 way. []

Ex. A personnel (HR) director needs to fill three distinct positions at a plant with selections from a group of 10 new engineers. The number of ways these positions can be filled is

Tn = 10 P 3 = 720 way. []

Ex. A manufacturer has 9 different motors, 2 of which came from the same supplier. These motors must be divided among 3 production lines with 3 motors going to each line. Once on a line the engines will be considered as indistinguishable. Find the probability that both motors from the particular supplier go to the first line.Well,

N = 9!/(3! 3! 3!) = 1680 way, and Tn = 7!/(3! 3!) = 140 (why?).So

P = 140/1680 = .0833. []

Lecture We now take up the important concept of the conditional probability (p. 50) of an event A occurring, given that event B has occurred. We write

P(A|B) = P(A ∩ B)/P(B) provided P(B) > 0.Conditional probability is essentially the idea that the prior occurrence of event B will possibly change the probability that A occurs. This is actually the case in most instances. For example, the probability that a certain team will run the ball late in the game is much higher if they are substantially ahead of the other team.

Ex. A great way to see conditional probability is through a contingency table. Below we have a contingency table for a marching band produced through the proc freq subroutine in SAS. This is done below for the χ square test (from Math 201).

Ex (cont)The SAS System The FREQ Procedure Table of Instrument by Gender

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Frequency‚ Expected male female Total

1 10 10 20Brass 8.5714 11.4292 5 15 20Wdwind 8.5714 11.4293 15 5 20Prcsn 8.5714 11.4294 0 10 10Front 4.2857 5.7143Total 30 40 70 Statistics for Table of Instrument by Gender Statistic DF Value Prob Chi-Square 0.0003 Sample Size = 70

Effectively the contingency table is a graphical representation of the (or sample) set which in this case has dim = 2. We can posit that the selections from this space are random and that we can define an r.v. function X: -> A ⊂ ℝ (although we will not take up the second step here). Note that

P(Male) = .4286, but P(Male|Brass) = .5. So the prior knowledge that we are selecting from the Brass players has an affect on the probability of selecting a male. []

Lecture In the previous example the effect of changing the probability of selecting a Male when our selection is restricted to Brass players is called dependence. Two events A and B are dependent if

P(A|B) ≠ P(A) (p. 52).Otherwise, we say that they are independent. Because of the mathematical definition of conditional probability, we have the following triangle of results.

P(A|B) = P(B), P(B|A) = P(B), P(A ∩ B) = P(A) P(B) (p. 52)The student should try to prove this using the result of the upcoming Theorem 2.5. Stemming from the conditional probability concepts, we have other probability laws. Theorem 2.5 tells us that a simple mathematical manipulation gives us that

P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B) (why ?).Further, we state without proof that

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (Theo. 2.6, p. 55). This last result is quite easy to see from the Venn and tree diagrams given earlier in the notes. Finally,

P(A) = 1 – P(Ac) (Theo. 2.7, p. 56).

Ex. Referring to the previous contingency table, we can see thatP(Prcsn|Male) = .5, but P(Male|Prcsn) = .75.

So we see that these are not necessarily identical unless the events have the exact same probability (why?). Ex (cont) Next observe that

P(Prcsn ∩ Male) = .2143

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= P(Male) x P(Prcsn|Male) = .4286 x .5.= P(Prcsn) x P(Male|Prcsn) = .2857 x .75.

So the rule works either way! []

Ex. In example 2.69 (p. 57) we have an instance where a volleyball player attempts to hit a ball on three tries. On each try she must alternate hands.

P(hit|right) = .7 and P(hit|left) = .4(which is actually pretty good). Victory is achieved if 2 hits are scored consecutively. Assuming independent trials, find the conditional probability that she wins given that she starts with her right hand.The solution can be found by constructing the sample space (Ω) in the form of a tree diagram for a right handed start seen below with probabilities in quantity marks. Observe the following diagram illustrating the process.

Right Left Right Final ProbabilityHit(.7) .7 x .4 x .7

Hit(.4)Miss(.3) .7 x .4 x .3

Hit(.7)Hit(.7)

Miss(.6)Miss(.3)

StartHit(.7) .3 x .4 x .7

Hit(.4)Miss(.3)

Miss(.3)Hit(.7)

Miss(.6)Miss(.3)

The winning combinations appear in Blue, while the irrelevant results appear in sky blue (notice the total of all probabilities at the ends of the tree diagram is 1). The probability that she wins (given a right handed start) is

P(win|RHstart) = .28 x 1.3 = .364.Further, the probability that she wins given that she gets a hit on the first try is given by

P(win|H1 with a RHstart) = .4. []

Ex. Problem 2.73 (p. 58) gives us an example of probability in electronics. We have a situation in which current flows from A to B if there is at least one closed path when the relays are activated.

Ex (cont) If the relays act independently, and close properly with probability .9 when activated, what is the probability that the current will flow (see below)?

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|---------a-----| |--------|A---> |---------b-----| |--------|-->---B

|---------c-----| |--------|P(proper flow) = 1 – P (all three fail)

= 1 – P(1 fails) P(2 fails) P(3 fails)= 1 - .13 = .999. P(c works| proper flow)= P(c works ∩ proper flow)/P(proper flow)= P(c works) x P(proper flow| c works)/P(proper flow)= .9 x 1/(.999) = .901. []

Ex. We are asked in problem 2.83 (p. 59) to find an example of an Ω set having events A, B, and C in which

P(A) > P(B) and P(C) > 0, but P(A|C) < P(B|C).Consider the example of the marching band given earlier, and let A be Percussion, B be Front and C be Female.

P(A) = .2857 > P(B) = .1429, and P(C) = .5714, butP(A|C) = .125 < P(B|C) = .25. []

Student Problem set AA1. Using the Gimp or MSPaint, draw a Venn diagram to illustrate that

a. (A ∪ B)c = Ac ∩ Bc.b. (A ∩ B)c = Ac ∪ Bc.

A2. Using the Gimp or MSPaint, draw a Venn diagram to illustrate thata. A = (A ∩ B) ∪ (A ∩ Bc).b. If B ⊂ A, then A = B ∪ (A ∩ Bc).

A3. Using the Gimp or MSPaint, draw a Tree diagram to illustrate the possible outcomes of the winning team in baseball's World Series given that team A has already won the first two games.A4. Using the Gimp or MSPaint, draw a Venn diagram to illustrate the possible outcomes for the winning team in the NFL playoffs.

A5. A survey is taken of 60 students at a university. 9 students lived off campus, 36 were undergraduates, and 3 were undergraduates living off campus. How many of these students were …

a) undergraduates or living off campus?

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b) undergraduates living on campus?c) graduate students living on campus?

A6. Suppose 3 patients (Miriam, Betty and Chris) walk into an outpatient clinic and randomly (to us) choose 1 of three available stations (Stn 1, Stn 2, Stn 3). If two or more are in the same station do not worry about order.

a. Write out the space for this experiment. How many outcomes are there?b. Let E be the event that each station has one patient. Identify the outcomes in

the Ω set satisfying this event.c. Find the probability that each station has one patient.

A7. Suppose 2 balanced dice are tossed and the total from each upper face is recorded. Let A be the event in which the second die face is even, and B be the event in which the total is even.

a. Write out the space for this experiment.b. Identify the subset A.c. Identify the subset B.d. Identify the subset A ∩ B, and find its probability.e. Identify the subset A B. and find its probability.

A8. If we wish to expand (x+y)8, what is the coefficient of x5y3? What is the coefficient of x3y5?A9. Probability played a role in the rigging of the April 24, 1980, Pennsylvania state lottery. To determine each digit of the three-digit winning number, each of the numbers 0,1,2,...,9 is placed on a ping pong ball, the ten balls are blown into a compartment, and the number selected for the digit is the one on the ball that floats to the top of the machine. Note that a ball may appear more than once. To alter the odds, the conspirators injected a liquid into all balls used in the game except those numbered 4 and 6, making it almost certain that the lighter balls would be selected and determine the digits in the winning number. Then they bought lottery tickets bearing the potential winning numbers. How many potential winning numbers were there (666 was the eventual winner)?A10. A group of three undergraduate and five graduate students are available to fill certain student government posts. If four students are to be randomly selected from this group, find the probability that exactly two undergraduates will be among the four chosen.A11. A balanced die is tossed six times and the number on the uppermost face is recorded each time. What is the probability that the numbers recorded are 1,2,3,4,5, and 6 in any order?A12. Five cards are dealt from a standard 52 card deck. What is the

a) P(3 aces and 2 kings (in no particular order))?b) P(3 cards of 1 kind and 2 of another kind (in no particular order))?

A13. Consider again the birthday problem from p. 3. What is the smallest value of n so that the probability (P) that at least two people share a birthday is ≥ .5?A14. Prove that (n+1 C k) = (n C k) + (n C k-1).

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Student Solution Set AA8. (8 C 5) = (8 C 3) = 56.A12. a. [(4 C 3)(4 C 2)]/(52 C 5) ≈ .000006.

b. We might want 3 hearts and 2 4s (P ≈ .0007). Or we might want 3 spades and 2 clubs (P ≈ .0086). (see output at right)A14. Try writing out the terms using factorials first. []

Lecture Definition 2.11 (p. 67) tells us that a partition of an Ω set is a set of mutually exclusive and exhaustive events. If the sets B1, B2, … Bn partition , then for any event A ∈ , Theorem 2.8 (p. 68) says that

P(A) = Σ nP(A|Bi)P(Bi).Of course, this is called the Law of Total Probability (LOTP).

Ex. Problem 2.86 (p. 65) gives us our first example of this concept. In a factory, 40% of the widgets are made at line 1, and 60% from line 2. These lines have defective rates of .08 and .10 respectively. What is the probability that a randomly selected item will not be defective? Clearly, the two lines make up the partition of Ω (the factory probability space ... set). Consequently, P(not defective)

= P(line 1) x P(not def.|line 1) + P(line 2) x P(not def.|line 2) = .4 x .92 + .6 x .9 = .908. []

Ex. The LOTP is powerful for proof. In problem 2.102 we are asked to show that if P(A|B) = P(A|Bc), then A und B are independent.

Note that the events B and Bc form a partition of . So P(A) = P(B)P(A|B) + P(Bc)P(A|Bc) (LOTP p. 68)= P(A|B) x (P(B) + P(Bc)) = P(A|B).

So A and B are independent (p. 52). //. []

Ex. Problem 2.110 (p. 71) says that under the “No pass, no play” rule, a HS has the following probabilities. If a student athlete has never been on probation, the probability of going on probation is .15. Ex (cont) But if the athlete is already on probation, the probability of coming off is .5. Under the current state,

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if 30% of athletes are already on probation, what is the probability that a randomly selected athlete will be on probation next semester? The partition of athletes is probation, non-probation. With that in mind the total probability for an athlete to be on probation next semester is

P(probation) = .3 x .5 + .7 x .15 = .255.So we could expect about 25% of the athletes to be on probation the next semester. []

Lecture Of course the whole reason for the LOTP is for Bayes's Rule. Using the LOTP we can assess the probability of an individual conditional event. Again, if the sets B1, B2, … Bn partition , and if P(Bi) > 0, for i = 1, 2, ... n, then for any event A ∈ , Theorem 2.9 (p. 68) says that

P(Bj|A) = P(A|Bj)P(Bj)/(Σ nP(A|Bi)P(Bi)),where j = 1, 2, ... n. Lecture The proof is based on the fact that

P(Bj|A) = P(A ∩ Bj)/P(A)in conjunction with the LOTP.

Ex Suppose that in the previous example an athlete on probation was, in fact, selected in the following semester. What is the probability that the athlete had been on probation the previous semester?Using Bayes's rule we have

P(Probation Semester 1 | Probation)= .3 x .5 / .255 = .5882.

So the probability is high that the athlete was on probation in the previous semester. []

Student Problem set BB1. A balanced die is cast 6 times and the number on the uppermost face is recorded each time. Find the probability that 1, 2, 3, 4, 5 and 6 are recorded in any order?B2. A survey of consumers in a community showed that 10% were dissatisfied with plumbing jobs done in their homes. Half the complaints dealt with plumber A, who does 40% of the plumbing jobs in the town.

a. Find the probability that a consumer will obtain an unsatisfactory plumbing job, given that the plumber was A.

b. Find the probability that a consumer will obtain a satisfactory plumbing job, given that the plumber was A.B3. A policy requiring all hospital employees to take a lie detector tests may reduce losses due to theft, but some employees regard such tests as a violation of their rights. Reporting on a hospital that used this procedure, the Orlando Sentinel Star noted that lie detectors have accuracy rates from 92% to 99%. B3 (cont) Looking into the risks employees face taking a lie detector test, suppose the probability is .05 that a particular detector indicates a person is lying, when they are telling the truth, and that any two tests are independent.

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a. What is the probability that a machine will conclude that each of three employees is lying when all are telling the truth?

b. What is the probability that the machine will conclude that at least one of the three employees is lying when all are telling the truth?B4. Three radar sets, operating independently, are set to detect any aircraft flying through a certain area. Each set has a probability of .02 of failing to detect a plane in its area.

a. If an aircraft enters the area, what is the probability that it goes undetected?

b. If an aircraft enters the area, what is probability that it is detected by all three radar sets?B5. A communications network has a built-in safeguard system against failures. In this system if line I fails, it is bypassed and line II is used. If line II also failes, it is bypassed and line III is used. The probability of failure of any one these three lines is .01, and the failures of these lines are independent events. What is the probability that this system of three lines does not completely fail?B6. Suppose that two balanced die are tossed repeatedly and the sum of the two uppermost faces is determined on each toss.

a. What is the probability that we obtain a sum of 3 before we obtain a sum of 7?b. What is the probability that we obtain a sum of 4 before we obtain a sum of 7?

B7. We know the following about a colormetric method used to test lake water for nitrates. If water specimens contain nitrates, a solution dropped into the water will cause the specimen to turn red 95% of the time. When used on water specimens without nitrates, the solution causes the water to turn red 10% of the time (because chemicals other than nitrates are sometimes present and they also react to the agent). Past experience in a lab indicates that nitrates are contained in 30% of the water specimens that are sent to the lab for testing.

a. If a water specimen is randomly selected from among those sent to the lab, what is the probability that it will turn red when tested?

b. If a water specimen is randomly selected and turns red when tested, what is the probability that it actually contain nitrates?B8. A plane is missing and is presumed to have equal probability of going down in any of three regions. If a plane is actually down in region i, let 1-αi denote the probability that the plane will be found upon a search of the ith region, i=1,2,3.

a. What is the conditional probability that the plane is in region 1 given that the search of region 1 was unsuccessful?

b. What is the conditional probability that the plane is in region 2 given that the search of region 1 was unsuccessful?

c. What is the conditional probability that the plane is in region 3 given that the search of region 1 was unsuccessful?B9. A spinner can land in any of four positions, A,B,C, and D, with equal probability. The spinner is used twice and the position is noted each time. Let the random variable Y

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denote the number of positions on which the spinner did not land. Compute the probabilities for each value of Y.B10. In the factory example on p 11, find P(Line 1 | defective) and P(Line 2 | def.).B11. Prove Bayes's Rule.

Student Solution Set BB1. Approximately .0154B2. P(A | D) = .5, P(A) = .4, P(D) = .1.

a) P(D | A) = P(A ∩ D)/P(A) = (P(D) P(A | D))/.4 = .1 * .5 / .4 = .05/.4 = .125.b) P(S | A) = 1 - .125 = .875.

B8. P(R1 | U1) = P(R1 ∩ U1)/P(U1) = (α1)/P(U1)? Note thatP(U1) = P(R1)P(U1|R1) + P(R2)P(U1|R2) + P(R3)P(U1|R3) (LOTP)

= .33(α1) + .67 (porque?).So P(R1 | U1) = .33(α1)/(.33(α1) + .67). []

Lecture Now it's time to take a closer look at a random variable that is discrete.

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