Homework 2. P(A 1 A 2 B) = P((A 1 A 2 ) B) P(B) From the distributive property of unions and intersections we have

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1 Homework 2 Readings: Sections 2.1, 2.2, 2.4, 2.5 A note: in Section 2.2, we have not talked yet about arbitrary sample spaces and the axioms of probability, instead we have focused on the case of a finite sample space, which corresponds to the part Determining Probabilities Systematically. Exercises: 1. Let A, B events with 0.5, 0.8, and P(A B) 0.9. Are A and B independent? Solution: From the addition rule we have P(A B) + P(A B) , and since , we conclude that P(A B), i.e., A and B are independent. 2. Addition rule for conditional probability If > 0, then show that for any events A 1 and A 2 we have P(A 1 A 2 B) P(A 1 B) + P(A 2 B) P(A 1 A 2 B). Remark: In a similar way you can extend other rules to conditional probabilities. For example, P(A B) + P(A c B) 1 for any event A. Solution: From the definition of conditional probability P(A 1 A 2 B) P((A 1 A 2 ) B). From the distributive property of unions and intersections we have (A 1 A 2 ) B (A 1 B) (A 2 B). Applying the addition rule in the numerator we obtain P((A 1 A 2 ) B) P((A 1 B) (A 2 B)) P(A 1 B) + P(A 2 B) P((A 1 B) (A 2 B)) P(A 1 B) + P(A 2 B) P((A 1 A 2 ) B) P(A 1 B) + P(A 2 B) P(A 1 A 2 B). 1

2 3. Implications of independence (a) Show that if A and B are independent events, then A c and B c are independent, i.e., P(A c B c ) P(A c ) P(B c ). Remark: In a similar way it can be shown that if A and B are independent, then A and B c are independent, A c and B are independent. Solution: We will show that A c, B c are independent. To this end, it suffices to show that P(A c B c ) P(A c ) P(B c ). From De Morgan s Laws we have that A c B c (A B) c, thus, P(A c B c ) P((A B) c ) 1 P(A B) 1 + P(A B) 1 + (1 ) (1 ) P(A c ) P(B c ), where the second equality follows from the complement rule, the third from the addition rule, the fourth from the definition of independence, and the last one again from the complement rule. (b) Show that if A, B, C are independent events, then A is independent of B c C c, i.e., P(A (B c C c )) P(B c C c ). Remark: In a similar way it can be shown that if A, B, C are independent, then A is independent of B C, of B C, of B \ C, of C \ B, etc. Solution: From De Morgan s Law we have B c C c (B C) c, thus, we need to show that P(A (B C) c ) P((B C) c ). Indeed, P(A (B C) c ) P(A \ (B C)) P(A B C) P(C) [1 P(C)] [1 P(B C)] P((B C) c ), 2

3 where in the second line we have applied the total probability law, the third line follows from the independence of A, B, C, and in the last line we have applied the complement rule. 4. A system of components (Ex. 80, Section 2.5, 8th Edition of the book) (a) When do we say that four events A 1, A 2, A 3, A 4 are independent? Hint: In class we defined the notion of independence for up to three events. Read the general definition of independence of n events in Section 2.5, and specialize it to the case n 4. (b) Consider the system of components connected as in the picture of the book (Ex. 80, Section 2.5, 8th Edition). Specifically, there are two subsystems connected in parallel. The first subsystem consists of components 1 and 2, which are connected in parallel. The second subsystem consists of components 3 and 4, which are connected in series. If components work independently of one another and each individual component fails with probability 10%, what is the probability that the system fails? (a) Solution: We say that A 1, A 2, A 3, A 4 are independent when the following hold and P(A i A j ) P(A i ) P(A j ) for every 1 i j 4, P(A i A j A k ) P(A i ) P(A j ) P(A k ) for every 1 i j k 4. and P(A 1 A 2 A 3 A 4 ) P(A 1 ) P(A 2 ) P(A 3 ) P(A 4 ). (b) Solution: Since the two sub-systems are connected in parallel, the system fails when both of them fail. The first sub-system fails when both components 1 and 2 fail. The second sub-system fails when at least one of components 3 and 4 fail. Let A i {component i fails}, i 1, 2, 3, 4. Then, {system fails} (A 1 A 2 ) (A 3 A 4 ) A 1 A 2 (A 3 A 4 ) The assumption that the components work independently of one another means that the events A 1, A 2, A 3, A 4 are independent (see part (a)). This implies that the events A 1, A 2, A 3 A 4 are independent (this can be checked in a similar way as in Ex. 3b). By the definition of independence we obtain P(system fails) P(A 1 A 2 (A 3 A 4 )) P(A 1 ) P(A 2 ) P(A 3 A 4 ). From the addition rule, we have P(A 3 A 4 ) P(A 3 ) + P(A 4 ) P(A 3 A 4 ) P(A 3 ) + P(A 4 ) P(A 3 ) P(A 4 ), 3

4 where in the second equality we have used the fact that A 3 and A 4 are independent, which is also implied by the independence of A 1, A 2, A 3, A 4 (see part (a)). Combining these two identities we obtain P(system fails) P(A 1 ) P(A 2 ) [P(A 3 ) + P(A 4 ) P(A 3 ) P(A 4 )] Since P(A i ) p 0.1 for every i 1, 2, 3, 4, P(system fails) p p (p + p p 2 ) p 3 (2 p) Missing aircrafts (Ex. 60, Section 2.4, 8th Edition of the book) 70% 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 a light aircraft has disappeared. (a) If it has an emergency locator, what is the probability that it will not be discovered? (b) If it does not have an emergency locator, what is the probability that it will be discovered? Solution: Consider the events D {missing aircraft is discovered} A {missing aircraft is equipped with an emergency locator}. We are given that P(D) 0.7, P(A D) 0.6, P(A c D c ) 0.9 From the complement rule for (conditional) probabilities (see Remark in Ex. 2), this implies that P(D c ) 0.3, P(A c D) 0.4, P(A D c ) 0.1. What we want to compute is P(D c A) in (a), and P(D A c ) in (b). In both cases, we have an application of Bayes rule. (a) From the definition of conditional probability we have P(D c A) P(Dc A) P(A Dc ) P(D c ). The denominator can be computed using the law of total probability as follows P(A D) + P(A D c ) P(A D) P(D) + P(A D c ) P(D c )

5 We conclude that P(D c A) P(A Dc ) P(D c ) % Therefore, if the aircraft has an emergency locator, the probability that it will not be discovered is roughly 6%. (b) Note that P(A c ) From the definition of conditional probability we have P(D A c ) P(D Ac ) P(A c ) P(Ac D) P(D) P(A c ) % Therefore, if the aircraft does not have an emergency locator, the probability that it will be discovered is roughly 51%. 6. Surveillance for terrorism (Ex. 67, Section 2.4, 8th Edition of the book) There has been a great deal of controversy over the last several years regarding what types of surveillance are appropriate to prevent terrorism. Suppose a particular surveillance system has a 99% chance of correctly identifying a future terrorist and a 99.9% chance of correctly identifying someone who is not a future terrorist. If there are 1000 future terrorists in a population of 300 million, and one of these 300 million is randomly selected, scrutinized by the system, and identified as a future terrorist, what is the probability that he/she actually is a future terrorist? Does the value of this probability make you uneasy about using the surveillance system? Explain. Solution: Let B represent the event that the selected person is identified as a terrorist, and A the event that the selected person is indeed a terrorist. We are given that P (B A).99, P(B c A c ).999, From Bayes rule we obtain P (B A) P(A B) P (B A) P (B A) + P (B A c ) P(A c ) ( ) Therefore, if a randomly selected person is scrutinized by the system and identified as a future terrorist, the probability that he/she actually is one is