Fall 2000. Homework #2: Solution. IE 230

Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers, John Wiley & Sons, New York, 1999. Chapter 3, Sections 3.1-3.2.

1. Consider the random experiment whose procedure is to randomly choose a student in this class. The outcome is the student's letter grade. Let the sample space be S = {A, B, C, F, I}, where I denotes "incomplete".

(a) Let E 1 = "the student passes the course". Let E 2 = "the student fails the course". Let E 3 = "the student does not pass the course". Write E 1, E 2, and E 3 in set notation by listing the relevant outcomes. ------E 1 = {A, B, C}, E 2 = {F}, and E 3 = {F, I}. ------

(b) Write the complement of E 1 in terms of E 2 and E 3. ------E 1′=E 3 or E 1′=E 2 ∪ E 3. ------

(c) Are E 1 and E 2 ∩ E 3 mutually exclusive? Argue why or why not by carefully using the de®nition. ------From Part (a), E 1 = {A, B, C} and E 2 ∩ E 3 = {F} ∩ {F, I} = {F}. Because they have no outcomes in common, E 1 ∩ (E 2 ∩ E 3) = ∅. Therefore, the sets E 1 and E 2 ∩ E 3 are mutually exclusive by de®nition. ------

(d) In words, what is E 2′? ------E 2′, the complement of E 2, is "the student does not fail the course". ------2. Venn Diagrams (a) With two Venn diagrams, illustrate the ®rst of DeMorgan's Laws: (A ∪ B)′=A′ ∩ B′. ------Draw two rectangles, each containing two overlapping circles labeled A and B. In the ®rst rectangle, shade the region not in A ∪ B. In the second rectangle, crosshatch (with a pattern of your choice) the region not in A and then (using a different pattern) the region not in B; then shade the region covered by both patterns. Notice that the shaded region in both rectangles is the same. ------(b) With a Venn diagram, illustrate A ∩ B ∩ C′. ------Draw a rectangle with three overlapping circles, labeled A, B, and C. Shade the region that is inside A and inside B and outside C. ------

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3. Consider an experiment where Purdue plays football games against Central Michigan, Kent State, Notre Dame, Minnesota, Northwestern, Penn Sate, Michigan, Wisconsin, Ohio State, Michigan State, and Indiana University, in that order. An outcome is the sequence of wins and losses.

Denote the outcome of an undefeated season by (W, W, W, W, W, W, W, W, W, W, W). More generally, denote an arbitrary outcome by (G 1, G 2,...,G 11), where Gi = W if Purdue wins game i and Gi = L if Purdue loses game i. The sample space is then {(G 1, G 2,...,G 11) | Gi = W or Gi = L} (a) Write the outcome in which Purdue loses to Central Michigan and then goes undefeated. ------(L, W, W, W, W, W, W, W, W, W, W) (An outcome is not a set, so no squiggly brackets.) ------(b) Write the event containing the two outcomes: an undefeated season or the outcome of Part (a). ------{(W, W, W, W, W, W, W, W, W, W, W), (L, W, W, W, W, W, W, W, W, W, W)} (An event is a set of outcomes, so squiggly brackets.) ------(c) In words, what is the event in Part (b)? ------"Purdue wins its last ten games". (Other answers could be correct.) ------(d) In words, what is the complement of the event in Part (b)? ------"Purdue does not win its last ten games" or "Purdue loses at least one of its last ten games". (Other answers could be correct.) ------(e) In the past, a college football game could end in a tie. Write that sample space? ------{(G 1, G 2,...,G 11) | Gi = W or Gi = L or Gi = T}, where T indicates a tie. ------4. If P(E) = 0.6, then (a) E is a(n) ___event______. (b) E will occur in about ______60_____ % of the replications. (c) P(E′) = __see below____. ------1 − P(E) = 1 − 0.6 = 0.4. ------

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5. (From Problem 3-24.) Pieces of sample foam from three suppliers are classi®ed for conformance to speci®cations. The results from a sample of 100 observations are summarized in the table. iiiiiiiiiiiiiiiconforms yes no 1182 supplier 2 17 3 35010

Let A denote the event that the piece is from supplier 1 and let B denote the event that the piece conforms to speci®cations. De®ne the experiment to be the (equally likely) choice of one of the 100 pieces of foam used to create the table. (a) What is the value of P(A)? ------P(A) = (18 + 2) / 100 = .20. ------(b) What is the value of P(A′B)? ------P(A′B) = (17 + 50) / 100 = .67. ------

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7. (From Problem 3-38.) Use the axioms of probability to show the following: (a) For any event E, P (E′) = 1 − P (E). ------Proof: 1 = P(S) Axiom 1. = P(E ∪ E′) Set theory. = P(E) + P(E′) Axiom 3.

Rearranging the terms yields the result. ------(b) P(∅) = 0. ------Proof: P(∅) = 1 − P(S) Complementary events (Part (a) above). = 1 − 1 Axiom 1. = 0 Arithmetic.

------(c) For any events A and B,ifA ⊂ B, then P (A) ≤ P (B). ------Proof: P(B) = P((B ∩ A) ∪ (B ∩ A′)) Set theory. = P(B ∩ A) + P(B ∩ A′) Axiom 3. = P(A) + P(B ∩ A′) A ⊂ B. ≥ P(A) Axiom 2.

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(d) If the events E 1, E 2, and E 3 are mutually exclusive, then P (E 1 ∪ E 2 ∪ E 3) = P (E 1) + P (E 2) + P (E 3). ------Proof:

P(E 1 ∪ E 2 ∪ E 3) = P((E 1 ∪ E 2) ∪ E 3) Set theory. = P(E 1 ∪ E 2) + P(E 3) Axiom 3. = P(E 1) + P(E 2) + P(E 3) Axiom 3.

(This logic extends directly to the union of n mutually exclusive events.) ------

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