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STAT E100 Section Week 5 - Probability

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Review - To best prepare for the exam, pay close attention to your homework and the solutions and take the practice midterms -Also, if you have questions about the grading of your assignment, feel free to contact your grader directly. Graders: Verbena, vkosovrasti@yahoo.comvkosovrasti@yahoo.com Ashley, amstilley@gmail.comamstilley@gmail.com

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Key Concepts: - Probability and Conditional Probability - Independence - Random Variables & Linear Transformations of Random Variables - Discrete and Continuous Random Variables

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Key Concepts: For any event A, 0 ≤ P(A) ≤ 1, Independent events: Disjoint or Mutually exclusive events: If A and B are disjoint, then P(A or B) = P(A) + P(B) Complement: Complement rule: P(A) = 1 - P(Ac) Remember: Not independent ≠ dependent

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SAMPLE QUESTION #1 1) Probability of Events I have 5 M&M’s left in the bag: 2 blue, 2 red, and 1 yellow. I plan on eating two more, and leaving the others in the bag. a) What is the sample space of outcomes for pairs of M&M’s that I will eat? b) Describe the set of events for (in terms of outcomes): A: eating two mismatched M&M’s (different colors). B: eating at least one blue M&M (they are my favorite). c) What is the probability of: A: B: d) Are A & B disjoint? ________ How do you know? e) Describe (in terms of outcomes), D, the intersection of A & B? and the union, E? f) What is the probability of D? of E? g) Are A & B Independent? How do you know?

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SAMPLE QUESTION #1 1) Probability of Events I have 5 M&M’s left in the bag: 2 blue, 2 red, and 1 yellow. I plan on eating two more, and leaving the others in the bag. a) What is the sample space of outcomes for pairs of M&M’s that I will eat? B= blue, Y= yellow, R= red Sample Space: BB, RR, YR, YB, BR, BY, RY, RB b) Describe the set of events for (in terms of outcomes): A: eating two mismatched M&M’s (different colors). Subspace A = {YR, YB, BR, BY, RY, RB} B: eating at least one blue M&M (they are my favorite). Subspace B = {BB, YB, BR, BY, RB} c) What is the probability of: A: 6/8 – See subspace A B: 5/8 – See subspace B d) Are A & B disjoint? ___No___ How do you know? There are outcomes that satisfy both events. e) Describe (in terms of outcomes), D, the intersection of A & B? and the union, E? D: eating two mismatched M&M’s and at least one blue M&M in the pair - Subset D = {YB, BR, BY, RB} E: eating two mismatched M&M’s or at least one blue M&M -Subset E = {BB, YR, YB, BR, BY, RY, RB} f) What is the probability of D? of E? P(D) = ½ P(E) = 7/8 P(E) = can also be calculated using => union = P(A) + P(B) – P(A and B)** = (6/8) + (5/8) – (1/2) = 7/8** **Be careful, since these events are not independent, P(A and B) does not equal P(A)*P(B), the calculation will give you 2 different answers. g) Are A & B Independent? How do you know? No, since knowing event A has occurred tells you something about the probability of event B occurring. Also, P(A)*P(B) does not equal P(A and B).

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SAMPLE QUESTION #2 2) Probability of Events I have 5 coins in my pockets: 2 quarters, 1 dime, 1 nickel, and 1 penny. I have 2 coins in my left pocket and the rest in my right, but I’m not sure which are where. a) What is the sample space for pairs of coins in my left pocket? b) Describe the set for events (in terms of outcomes): A: getting a sum of the two coins to be even? B: of getting at least one quarter. c) What is the probability of: A: B: d) Are A & B disjoint? ________ How do you know? e) Describe, D, the intersection of A & B? How about the union, E? f) What is the probability of D? of E? g) Are A & B Independent? How do you know?

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SAMPLE QUESTION #2 2) Probability of Events I have 5 coins in my pockets: 2 quarters, 1 dime, 1 nickel, and 1 penny. I have 2 coins in my left pocket and the rest in my right, but I’m not sure which are where. a) What is the sample space for pairs of coins in my left pocket? Q = quarter, D = dime, N = nickel, P = penny Sample Space: QQ, QD, QN, QP, DN, DP, NP b) Describe the set for events (in terms of outcomes): A: getting a sum of the two coins to be even? Subspace A = QQ, QN, QP, NP B: of getting at least one quarter. Subspace B = QQ, QD, QN, QP c) What is the probability of: A: 4/7 – See Subspace A B: 4/7 – See Subspace B d) Are A & B disjoint? ___No___ How do you know? There are outcomes that satisfy both events. e) Describe, D, the intersection of A & B? How about the union, E? D: getting a sum of the two coins to be even and getting at least one quarter: Subspace D = {QQ, QN, QP} E: getting a sum of the two coins to be even or getting at least one quarter: Subspace E = {QQ, QD, QN, QP, NP} f) What is the probability of D? of E? P(D) = 3/7 P(E) = 5/7 g) Are A & B Independent? How do you know? No, since knowing event A has occurred tells you something about the probability of event B occurring. Also, P(A)*P(B) does not equal P(A and B).

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SAMPLE QUESTION #3 Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have such a locator. Suppose that a light aircraft has disappeared. If it has an emergency locator, what is the probability that it will be discovered? Let D = event light aircraft is discovered if and when it disappears when in flight and E = event aircraft has an emergency locator From http://sites.stat.psu.edu/~lsimon/stat250/homework/chapter3/bayes.pdf

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SAMPLE QUESTION #3 Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas 90% of the aircraft not discovered do not have such a locator. Suppose that a light aircraft has disappeared. If it has an emergency locator, what is the probability that it will be discovered? Let D = event light aircraft is discovered if and when it disappears when in flight and E = event aircraft has an emergency locator From http://sites.stat.psu.edu/~lsimon/stat250/homework/chapter3/bayes.pdf Given: P(D) = 0.70 P(D c ) = 1- P(D) = 1 - 0.70 = 0.30 P(E|D) = 0.60 P(E c |D c ) = 0.90 = P(E|D c ) = 1 - P(E c |D c ) = 1 - 0.90 = 0.10 Then, P(D|E) = P(D and E)/P(E) = P(D and E)/P((E and D) OR (E and D c )) = P(D and E)/{P(E and D) + P(E and D c )} = {P(D) × P(E|D)}/{(P(D) × P(E|D)) + (P(D c ) × P(E|D c ))} = (0.70 × 0.60)/((0.70×0.60) + (0.30 × 0.10)) = 0.42/(0.42+0.03) = 0.42/0.45 = 0.93

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