Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers, John Wiley & Sons, New York, 1999. Sections 3.6-3.7. Much of the Concise Notes page 5. Throughout, use the step-by-step approach whenever possible. 1. (From M&R, Problem 3-73.) In a test of a printed circuit board using a random test pattern, each bit in an array of ten bits is equally likely to be a zero or a one. Assume that the bits are independent. Let Z i denote the event that the i th bit is a zero, i = 1, 2,...,. (a) Describe the experiment and a sample space S so that the events Z 1, Z 2,...,Z are meaningful. Experiment: Randomly choose one printed circuit board. Sample space: The set of all available printed circuit boards. The events Z i are the circuit boards in S for which bit i is zero. (b) Write the event "all ten bits are zero". Z 1 Z 2... Z or equivalently (c) Determine the probability of the event in Part (b). P( Z i ) = Π P(Z i ) Z i since independent = (1 / 2) since P(Z i ) = 1 / 2 = 0.000977 simplify (d) Show that the probability of "alternating bits" (that is, a zero always following a one and a one always following a zero) is twice as likely as all bits being zero. P( alternating bits ) = P[(Z 1 Z 2 Z 3... Z ) (Z 1 Z 2 Z 3... Z )] = P(Z 1 Z 2 Z 3... Z ) same event + P(Z 1 Z 2 Z 3... Z ) mut. excl. = P(Z 1 )(Z 2 )P(Z 3 )... P(Z )) + P(Z 1 )P(Z 2 )P(Z 3 )... P(Z ) independence = (1 / 2) + (1 / 2) P(Z i ) = 1 / 2 = 2P( Z i ) from Part (b) - 1 of 6 - Schmeiser
2. (From M&R, Example 3-28) A random circuit, composed of a network of n devices, operates if and only if there is a path from beginning to end. The path cannot pass through a device that is not working. (Think of electricity flowing.) Let D i denote the event that device i works, i = 1, 2,..., n. Assume that these events are jointly independent. Consider the network of n = 6 devices illustrated in the textbook. For the network to operate, at least one of the first three devices must work, at least one of the fourth and fifth devices must work, and the sixth device must work. (a) Using D i event notation, write the six device probabilities given in the textbook. P(D 1 ) = P(D 2 ) = P(D 3 ) = 0.9 P(D 4 ) = P(D 5 ) = 0.95 P(D 6 ) = 0.99 (b) Write, in set notation, the event that the network operates. (D 1 ) (D 4 ) D 6 (c) Compute the probability of the event in Part (b). P(D 1 ) (D 4 ) D 6 = P(D 1 )P(D 4 )P(D 6 ) independence = [1 P((D 1 ) )] [1 P((D 4 ) )] P(D 6 ) complements = [1 P(D 1 D 2 D 3 )] [1 P(D 4 D 5 )] P(D 6 ) DeMorgan s Law = [1 P(D 1 )P(D 2 )P(D 3 )] [1 P(D 4 )P(D 5 )] P(D 6 ) independence = [1 (1 P(D 1 )) (1 P(D 2 )) (1 P(D 3 )] [1 (1 P(D 4 )) (1 P(D 5 ))] P(D 6 ) complements = [1 (1 0.90) (1 0.90) (1 0.90)] substitute information [1 (1 0.95) (1 0.95)] (0.99) from Part (b) = (1 0.1 3 )(1 0.05 2 ) (0.99) simplify = 0.987 simplify - 2 of 6 - Schmeiser
3. An unknown person has won the $1,000,000 lottery. To avoid greedy acquaintances, the person tries to claim the prize in secret, but with only partial success. Only four people have been seen entering the lottery office: a short man named Sam, a tall man named Ted, a short woman name Sue, and a tall woman named Tia; one is the winner. Having no other information, assume that each is equally likely to have won. We begin an investigation. Let M denote the event that the winner is a man. Let T denote the event that the winner is tall. Let U denote the event that the winner is either a tall woman or a short man. (a) Show that M and T are independent. M andt are independent if and only if P(M T ) = P(M )P(T). All four people are equally likely, so all three probabilities are the fraction of the four people in the event. Therefore P(M T ) = P("Ted" ) = 1 / 4. P(M) = P("Ted" or "Sam" ) = 2 / 4. P(T) = P("Ted" or "Tia" ) = 2 / 4. Therefore M and T are independent. (b) Show that three events M, T, and U are pairwise independent. Three events are pairwise independent if each of all three pairs of events are independent. From Part (a) M and T are independent. Similarly, P(M U ) = P("Sam" ) = 1 / 4. P(M) = P("Ted" or "Sam" ) = 2 / 4. P(U) = P("Tia" or "Sam" ) = 2 / 4. implies that M and U are independent. Similarly, P(T U ) = P("Tia" ) = 1 / 4. P(T) = P("Ted" or "Tia" ) = 2 / 4. P(U) = P("Tia" or "Sam" ) = 2 / 4. implies that T and U are independent. (c) Use the definition to show that the three events M, T, and U are not jointly independent. To be jointly independent, the three events M, T, and U must be pairwise independent and P(M T U ) = P(M )P(T)P(U) Again because all four people are equally likely to have won, the probabilities are the fraction of people in the event. Therefore, P(M T U ) = P( ) = 0, which is not equal to P(M )P(T) P(U) = 1 / 2 3, which implies that M, T, and U are not jointly independent. - 3 of 6 - Schmeiser
(d) Suppose that a lottery guard knows that the winner is a man and that a lottery clerk knows that the winner is tall. Compute P(U M T ). Explain in words why your answer makes sense. P(U M T ) = P(U (M T )) definition of cond. probability P(M T ) = 0 / (1 / 4) substituting from Parts (c) and (c) = 0 simplifying (e) Roughly, event independence means that knowing whether an event occurs is not useful for predicting whether another event occurs (in the same replication of the experiment). Therefore, because of pairwise independence, M helps to predict neither T nor U. Because the three events are not jointly independent, something can be predicted. What? As shown in Part (d), knowing that the winner is a tall man (i.e. T M ) is sufficient information to eliminate Tia and Sam (i.e., U ). This one example is enough to demonstrate that the events are not jointly independent. Other examples are easy to find. Comment. Compare this problem to Problem 2 from Fall 2000. 4. Result: If the events A and B are independent, then A and B are independent. Provide the reason for each step of the proof. P(A B ) = P((A B ) ) < DeMorgan s Law > = 1 P(A B ) < complement > = 1 [P(A ) + P(B ) P(A B )] < maybe not m.e. > = 1 [P(A ) + P(B ) P(A )P(B )] < independence > = [1 P(A )] [1 P(B )] < algebra > = P(A )P(B ) < complements > - 4 of 6 - Schmeiser
5. In a trial by jury, a mistake can occur in two ways. First, the jury can convict an innocent defendant. Second, the jury can free a guilty defendant. Let G denote the event that the defendant is guilty. Let F denote the event that the jury frees the defendant. Consider this situation. A robbery victim views a line up of seven suspects, one of whom is guilty. The victim, embarrassed by not recognizing the guilty person, randomly chooses one of the seven, who is then tried. Assume that a jury convicts a guilty defendant 95% of the time and frees an innocent defendant 99% of the time. (a) Write the given information as probabilities of events. P(G ) = 1 / 7 since equally likely P(F G ) = 0.95 P(F G ) = 0.99 (b) Are events G and F independent? Why or why not? G and F are not independent. If freedom were not dependent upon guilt, the court system would be worthless. One way to show the dependence is to notice that 5% of guilty defendants are freed (P(F G ) = 1 P(F G ) = 1 0.95 = 0.05). The corresponding probability for innocent defendants is P(F G ) = 0.99. Because the probability of F depends upon G, they are not independent. (c) Conditional that the defendant is convicted, what is the probability that he or she is innocent? P(G F ) = = P(F G )P(G ) P(F G )P(G ) + P(F G )P(G) [1 P(F G )] [1 P(G )] [1 P(F G )] [1 P(G )] + P(F G )P(G) Bayes s Theorem complements = [1 (0.99)] [1 (1 / 7)] [1 (0.99)] [1 (1 / 7)] + (0.95) (1 / 7) complements = (0.01) (6 / 7) (0.01) (6 / 7) + (0.95) (1 / 7) from Part (a) = 0.00857 0.00857 + 0.136 simplify = 0.059 simplify - 5 of 6 - Schmeiser
6. Spread sheet. (Email your spreadsheet to "ie230@ecn.purdue.edu". The results from Part (g) should be handwritten and submitted with your solutions to Problems 1-5.) We will perform a Monte Carlo simulation, at first with one-hundred independent replications, to estimate the probability that the network of Problem 2 operates. One purpose of such a simulation is to check your answer. For more-complicated networks, however, the simulation might be the only way to analyze the network. (a) At the top of the sheet, place your name, class, homework number, and problem number. Leave four or five blank rows. (b) In Column A place the heading "Trial #". Beneath the heading place the integers 1 through 0. (You can "drag" the integers; you should not enter the numbers manually.) (c) In Columns B-G place the headings U1, U2,..., U6. Beneath each of these headings, place 0 random numbers. (Use the "rand" command.) (d) In Columns H-M place the headings D1, D2,..., D6. Above each of these heading place the corresponding given probability of the device operating. Beneath each of these six headings, place 0 binary numbers, 1 if D i occurred (that is, if device i worked) and 0 if D i did not occur (that is, if device i did not work). (Use the "if" command, just like flipping a biased coin. The results for Column H should use the random numbers in Column B, I from C, etc.) (e) In Column N place the heading "Operates?" Beneath this heading compute the 0 binary numbers that indicate whether the network operated on each replication. As before, let 1 indicate that the event occurred and 0 that it did not occur. (You can use the "if" function. But it is easier to use the expression [1 (1 H 6)(1 I 6)(1 J 6)] [1 (1 K 6)(1 L 6)] M 6 for the trial in row 6. ) (f) Above the Column N heading, compute the relative frequency that the network operated. (Use the "average" command.) Label this cell. (g) Hit F9 enough times to convince yourself that the results match your answer from Problem 2(c). If the results do not match, check your work. When they do match, write the relative frequencies obtained from hitting F9 ten times. See the MSExcel spreadsheet for the solution to Problem #6. - 6 of 6 - Schmeiser