Chapter 2. Probability. 2.1 Basic ideas of probability


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1 Chapter 2 Probability 2.1 Basic ideas of probability One of the most basic concepts in probability and statistics is that of a random experiment. Although a more precise definition is possible, we will restrict ourselves here to understanding a random experiment as a procedure which is carried out under a certain set of conditions; it can be repeated any number of times under the same set of conditions, and upon the completion of the procedure certain results are observed. The results obtained are denoted by s and are called sample points. The set of all possible sample points is denoted by S and is called a sample space. Subsets of S are called events and are denoted by capital letters A, B, C, etc. An event consisting of one sample point only, {s},is called a simple event and composite otherwise. An event A occurs (or happens if the outcome of the random experiment (that is, the sample point s belongs in A, s A; A does not occur (or does not happen if s / A. The event S always occurs and is called the sure or certain event. On the other hand, the event never happens and is called the impossible event. The complement of the event A, denoted by A c, is the event defined by: A c = {s S; s / A}. So A c occurs whenever A does not, and vice versa. The union of the events A 1,...,A n, denoted by A 1... A n or n j=1 A j, is the event defined by n j=1 A j = {s S; s A j, for at least one j = 1,...,n}. So the event n j=1 A j occurs whenever at least one of A j,j = 1,...,n occurs. The 1
2 CHAPTER 2. PROBABILITY 2 definition extends to an infinite number of events. Thus, for countably infinite many events A j,j = 1, 2,..., one has j=1 A j = {s S; s A j, for at least one j = 1, 2,...}. The intersection of the events A j, j = 1,...,n is the event denoted by A 1... A n or n j=1 A j and is defined by n j=1 A j = {s S; s A j, for all j = 1,...,n}. Thus, n j=1 A j occurs whenever all A j, j = 1,...,n occur simultaneously. This definition extends to an infinite number of events. Thus, for countably infinite many events A j,j = 1, 2,..., one has j=1 A j = {s S; s A j, for all j = 1, 2,...}. If A 1 A 2 =, the events A 1 and A 2 are called disjoint. The events A j, j = 1, 2,..., are said to be mutually or pairwise disjoint, if A i A j = whenever i j.
3 CHAPTER 2. PROBABILITY 3 The differences A 1 A 2 and A 2 A 1 are the events defined by A 1 A 2 = {s S; s A 1,s / A 2 }, A 2 A 1 = {s S; s A 2,s A 1 }. From the definition of the preceding operations, the following properties follow immediately 1. S c =, c = S, (A c c = A. 2. S A = S, A = A, A A c = S, A A = A. 3. S A = A, A =, A A c =, A A = A. 4. Commutative laws A 1 A 2 = A 2 A 1 A 1 A 2 = A 2 A 1 5. Associative laws A 1 (A 2 A 3 = (A 1 A 2 A 3 A 1 (A 2 A 3 = (A 1 A 2 A 3 6. An identity: j A j = A 1 (A c 1 A 2 (A c 1 A c 2 A Distributive laws A ( j A j = j (A A j A ( j A j = j (A A j 8. DeMorgan s laws: ( j A j c = j A c j ( j A j c = j A c j In the last relations, when the range of the index j is not indicated explicitly, it is assumed to be a finite set, such as 1,...,n, or a countably infinite set, such as 1, 2,... Formally, a random variable, to be shortened to r.v., is simply a function defined on a sample space S and taking values in the real line R = (,. Random variables are denoted by capital letters, such as X, Y, Z, with or without
4 CHAPTER 2. PROBABILITY 4 subscripts. Thus, the value of the r.v. X at the sample point s is X(s, and the set of all values of X, that is, the range of X, is usually denoted by X(S. Two kinds of r.v. s emerge: discrete r.v. s (or r.v. s of the discrete type, and continuous r.v. s (or r.v. s of the continuous type. A r.v. X is called discrete, if X takes on countably many values; i.e., either finitely many values such as x 1,...,x n, or countably in infinite many values such as x 1,x 2,... On the other hand, X is called continuous (or of the continuous type if X takes all values in a proper interval I R. Example 2.1 Let S = {(x,y R 2 ; 3 x 3, 0 y 4, x and y integers} and define r.v. X by: X((x,y = x + y. Determine the values of X, as well as the following events: (X 2, (3 < X 5, (X > 6. X takes on the values: 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7. (X 2 ={ (3, 0, (3, 1, (3, 2, (3, 3, (3, 4, (2, 0, (2, 1, (2, 2, (2, 3, (2, 4, (1, 0, (1, 1, (1, 2, (1, 3, (0, 0, (0, 1, (0, 2, (1, 0, (1, 1, (2, 0 } (3 < X 5 = (4 X 5 = (X = 4 or X = 5 = { (0, 4, (1, 3, (1, 4, (2, 2, (2, 3, (3, 1, (3, 2 } (X > 6 = (X 7 = { (3, 4 } Probability is a function, denoted by P, defined for each event of a sample space S, taking on values in the real line R, and satisfying the following three properties: (P1 P(A 0 for every event A (nonnegativity of P. (P2 P(S = 1 (P is normed. (P3 For countably infinite many pairwise disjoint events A i, i = 1, 2,..., A i Aj =, i j, it holds ( P A i = P(A i Next, we present some basic results following immediately from the defining properties of the probability: 1. For an empty set P( = 0 From the obvious fact that S = S... and property (P3, P(S = P(S... = P(S + P( + P( +... or P( + P( +... = 0. By (P1, this can only happen when P( = 0.
5 CHAPTER 2. PROBABILITY 5 2. For any pairwise disjoint events A 1,...,A n, P( n A i = n P(A i. Take A i = for i n + 1, consider the following obvious relation, and use (P3 and #1 to obtain: ( n ( n P A i = P A i = P(A 1 = P(A i. 3. For any event A, P(A c = 1 P(A. From (P2 and #2, P(A A c = P(S = 1 or P(A + P(A c = 1, so that P(A c = 1 P(A. 4. A 1 A 2 implies P(A 1 P(A 2 and P(A 2 A 1 = P(A 2 P(A 1. The relation A 1 A 2, implies A 2 = A 1 (A 2 A 1, so that, by #2, P(A 2 = P(A 1 + P(A 2 A 1. Solving for P(A 2 A 1, we obtain P(A 2 A 1 = P(A 2 P(A 1, so that, by (P1, P(A 1 P(A 2. At this point it must be pointed out that P(A 2 A 1 need not be P(A 2 P(A 1, if A 1 is not contained in A P(A 1 for every event A. Clearly, A S for any event A. Then (P1, #1 and #4 give: 0 = P( P(A P(S = (a For any two events A 1 and A 2 : P(A 1 A 2 = P(A 1 + P(A 2 P(A 1 A 2. It is clear (by means of a Venn diagram, for example that A 1 A 2 = A 1 (A 2 A c 1 = A 1 (A 2 A 1 A 2.
6 CHAPTER 2. PROBABILITY 6 Then, by means of #2 and #4: P(A 1 A 2 = P(A 1 + P(A 2 A1 A 2 = P(A 1 + P(A 2 P(A 1 A (b For any three events A 1, A 2, and A 3 : P(A 1 A 2 A 3 = P(A 1 + P(A 2 + P(A 3 (P(A 1 A 2 +P(A 1 A 3 + P(A 2 A 3 + P(A 1 A 2 A 3. Apply part (a to obtain: P(A 1 A 2 A 3 = P [(A 1 A 2 A 3 ] = P(A 1 A 2 + P(A 3 P [(A 1 A 2 A 3 ] = P(A 1 + P(A 2 P(A 1 A 2 + P(A 3 P [(A 1 A 3 (A 2 A 3 ] = P(A 1 + P(A 2 + P(A 3 P(A 1 A 2 [P(A 1 A 3 + P(A 2 A 3 P(A 1 A 2 A 3 ] = P(A 1 +P(A 2 +P(A 3 P(A 1 A 2 P(A 1 A 3 P(A 2 A 3 +P(A 1 A 2 A For any events A 1, A 2,..., P( A i P(A i, and P( n A i n P(A i. By the identity given in the start of the section and (P3: ( P A i = P [ A 1 (A c 1 A 2... ( ] A c 1... A c n 1 A n...
7 CHAPTER 2. PROBABILITY 7 = P(A 1 + P (A c 1 A P ( A c 1... A c n 1 A n +... P(A 1 + P(A P(A n +... (by #4. For the finite case: ( n P A i = P [ A 1 (A c 1 A 2... ( ] A c 1... A c n 1 A n = P(A 1 + P(A c 1 A P ( A c 1... A c n 1 A n P(A 1 + P(A P(A n. Generalization of property #6 to more than three events is given as a theorem below: Theorem 2.1 The probability of the union of any n events, A 1,...,A n, is given by: ( n P A j = j=1 n P(A j P(A j1 A j2 j=1 1 j 1 <j 2 n + P(A j1 A j2 A j j 1 <j 2 <j 3 n +( 1 n+1 P(A 1... A n. First, sum up the probabilities of the individual events, then subtract the probabilities of the intersections of the events, taken two at a time (in the ascending order of indices, then add the probabilities of the intersections of the events, taken three at a time as before, and continue like this until you add or subtract (depending on n the probability of the intersection of all n events.
8 CHAPTER 2. PROBABILITY 8 Example 2.2 Students in a certain college subscribe to three news magazines A, B, and C according to the following proportions: A: 20%, B : 15%, C : 10%, both A and B : 5%, both A and C : 4%, both B and C : 3%, all three A, B, and C : 2%. If a student is chosen at random, what is the probability he/she subscribes to none of the news magazines? P(A c B c C c = P((A B C c = 1 P(A B C Answer: 0.65 = 1 [P(A + P(B + P(C P(A B P(A C P(B C + P(A B C] = 1 ( = = 0.65 We first briefly describe two somewhat classical methods for assigning probabilities to random variables: the equal likelihood model and the relative frequency method. When we have an experiment where each of n outcomes is equally likely, then we assign a probability mass 1/n of to each outcome. When the equal likelihood assumption is not valid, then the relative frequency method can be used. With this technique, we conduct the experiment n times and record the outcome. The probability of event E is assigned by P(E = f/n, where f denotes the number of experimental outcomes that satisfy event E. Example 2.3 Consider a wellshuffled deck of 52 cards, and suppose we draw at random three cards. What is the probability that at least one is an ace? Let A be the required event, and let A i be defined by: A i = exactly i cards are aces, i = 0, 1, 2, 3. Then, P(A = P(A 1 A 2 A 3 = 1 P(A 0 P(A 0 = = 4, 324 5, 525 so that P(A = 1,201 5,525 =
9 CHAPTER 2. PROBABILITY 9 Answer: Another way to find the desired probability that an event occurs is to use a probability density function (pdf which will be discussed in Chapters 4 and 5. Exercises Exercise 2.1 In terms of the events A 1, A 2, A 3 in a sample space S and their complements, express the following events: 1. B 0 = {s S; s belongs to none of A 1, A 2, A 3 }, 2. B 1 = {s S; s belongs to exactly one of A 1, A 2, A 3 }, 3. B 2 = {s S; s belongs to exactly two of A 1, A 2, A 3 }, 4. B 3 = {s S; s belongs to all of A 1, A 2, A 3 }, 5. C = {s S; s belongs to at most two of A 1, A 2, A 3 }, 6. D = {s S; s belongs to at least one of A 1, A 2, A 3 }. 1. B 0 = A c 1 A c 2 A c 3 2. B 1 = (A 1 A c 2 A c 3 (A c 1 A 2 A c 3 (A c 1 A 2 c A 3 3. B 2 = (A 1 A 2 A c 3 (A 1 A c 2 A 3 (A c 1 A 2 A 3 4. B 3 = A 1 A 2 A 3 5. C = B 0 B 1 B 2 6. D = B 1 B 2 B 3 = A 1 A 2 A 3 Exercise 2.2 If the events A, B, and C are related as follows: A B C and P(A = 1, 4 P(B = 5, and P(C = 7, compute the probabilities of the following events: A c B
10 CHAPTER 2. PROBABILITY A B c C c 3. A c B c C c 1. A c B = B A c = B A and A B. Therefore P(A c B = P(B A = P(B P(A = = 1 6 = A B c C c = A (B c C c = A (B C c = A C c = A C =, so that P(A B c C c = A c B c C c = (A B C c = C c, so that P(A c B c C c = P(C c = 1 P(C = = 5 12 = Answer: 0.167; 0; Exercise 2.3 Let A and B be the respective events that two contracts I and II, say, are completed by certain deadlines, and suppose that: P(at least one contract is completed by its deadline = 0.9 and P( both contracts are completed by their deadlines = 0.5. Calculate the probability: P(exactly one contract is completed by its deadline. We have P(A B = 0.9 and P(A B = 0.5. We need to calculate P((A B c (A c B = P(A B c + P(A c B. Clearly, A = (A B (A B c and B = (A B (A c B, so that P(A = P(A B + P(A B c and P(B = P(A B + P(A c B. Hence, P(A B c = P(A P(A B and P(A c B = P(B P(A B. Then P(A B c + P(A c B = P(A + P(B 2P(A B = [P(A + P(B P(A B] P(A B = P(A B P(A B = = 0.4. Answer: 0.4 Exercise 2.4 A foursided die has the numbers 1 through 4 written on its sides, one on each side. If the die is rolled twice: 1. Write out a suitable sample space S.
11 CHAPTER 2. PROBABILITY If X is the r.v. denoting the sum of numbers appearing, determine the values of X. 3. Determine the events: X 3, 2 X < 5, X > S = { (1, 1, (1, 2, (1, 3, (1, 4, (2, 1, (2, 2, (2, 3, (2, 4, (3, 1, (3, 2, (3, 3, (3, 4, (4, 1, (4, 2, (4, 3, (4, 4 } 2. The values of X are: 2, 3, 4, 5, 6, 7, (X 3 = (X = 2 or X = 3 = {(1, 1, (1, 2, (2, 1}, (2 X < 5 = (2 X 4 = (X = 2 or X = 3 or X = 4 = {(1, 1, (1, 2, (2, 1, (1, 3, (2, 2, (3, 1} (X > 8 = 2.2 Conditional probability Conditional probability is a probability in its own right, and it is an extremely useful tool in calculating probabilities. Essentially, it amounts to suitably modifying a sample space S, associated with a random experiment, on the evidence that a certain event has occurred. Example 2.4 Tossing three distinct coins once. Then, with H and T standing for heads and tails respectively, a sample space is: S = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}. Consider the events B = exactly 2 heads occur = {HHT, HTH, THH}, A = coins #1 and #2 show heads = {HHH, HHT }. Then P(B = 3 and P(A = 8 2 = 1. Now, suppose we are told that event B has occurred and we are asked to 8 4 evaluate the probability of A on the basis of this evidence. Clearly, what really matters here is the event B, and, given that B has occurred, the event A occurs only if the sample point HHT appeared; that is, the event {HHT } = A B occurred. The required probability is then 1 = 1/8 = P(A B > 3 3/8 P(B 1 = P(A. 4
12 CHAPTER 2. PROBABILITY 12 The conditional probability of an event A, given the event B with P(B > 0, is denoted by P(A B and is defined by: P(A B = P(A B/P(B. Replacing B by the entire sample space S, we are led back to the (unconditional probability of A, as P(A S = P(A. Thus, the conditional probability is a P(S generalization of the concept of probability where S is restricted to an event B. The conditional probability is a probability can be seen formally as follows: P(A B 0 for every A by definition; P(S B = P(S B P(B = P(B P(B = 1; and if A 1, A 2,... are pairwise disjoint, then ( P A j B = P [( ] j=1a j B P(B j=1 j=1 = P(A j B = P(B = P [ j=1(a j B ] P(B j=1 P(A j B P(B = P(A j B. It is to be noticed, furthermore, that the P(A B can be smaller or larger than the P(A, or equal to the P(A. The case that P(A B = P(A is of special interest. If P(A B = P(A then the occurrence of event B provides no information in reevaluating the probability of A. Under such a circumstance, it is only fitting to say that A is independent of B. If, in addition, P(A > 0, then B is also independent of A because j=1 P(B A = P(B A P(A = P(A B P(A = P(A BP(B P(A = P(AP(B P(A = P(B. Because of this symmetry, we then say that A and B are independent. From the definition of either P(A B or P(B A, it follows then that P(A B = P(AP(B. Two events A 1 and A 2 are said to be independent (statistically or in the probability sense, if P(A 1 A 2 = P(A 1 P(A 2. When P(A 1 A 2 P(A1P(A2 they are said to be dependent. The definition of independence can be extended to any number n of events A 1,...,A n by requiring any combination of the events to be independent. The events A 1,...,A n are said to be independent (statistically or in the probability sense if, for all possible choices of k out of n events (2 k n, the probability of their intersection equals the product of their probabilities. More formally, for any k with 2 k n and any integers j 1,...,j k with 1 j 1 < < j k n, we have: ( k k P A ji = P(A ji. Example 2.5
13 CHAPTER 2. PROBABILITY 13 Let S = {1, 2, 3, 4} and let P({1} = P({2} = P({3} = P({4} = 1/4. Define the events A 1, A 2, A 3 by: A 1 = {1, 2},A 2 = {1, 3},A 3 = {1, 4}. Check if events A 1, A 2 and A 3 are independent. P(A 1 = P(A 2 = P(A 3 = 1 2 P(A 1 A 2 = P({1} = 1 4 = P(A 1 P(A 2, P(A 1 A 3 = P({1} = 1 4 = P(A 1 P(A 3, P(A 2 A 3 = P({1} = 1 4 = P(A 2 P(A 3. However, P(A 1 A 2 A 3 = P({1} = 1 4 P(A 1P(A 2 P(A 3 = 1 8. Events are dependent. We can calculate the probability of the intersection of n events, step by step, by means of conditional probabilities using following theorem: Theorem 2.2 Multiplicative Theorem For any n events A 1,..., A n with P( n 1 j=1 A j > 0, it holds: ( n P A j = P(A n A 1... A n 1 P(A n 1 A 1... A n 2 j=1...p(a 2 A 1 P(A 1. For n = 2, the theorem is true since p(a 2 A 1 = P(A 1 P 2 P(A 1 yields P(A 1 A 2 = P(A 2 A 1 P(A 1. Next, assume P(A 1... A k = P(A k A 1... A k...p(a 2 A 1 P(A 1 and show that P(A 1...A k+1 = P(A k+1 A 1... A k P(A k A 1... A k 1...P(A 2 A 1 p(a 1. Indeed, P(A 1... A k+1 = P((A 1... A k A k+1 = P(A k+1 A 1... A k P(A 1... A k = P(A k+1 A 1... A k P(A k A 1... A k 1...P(A 2 A 1 P(A 1 by the induction. Example 2.6 An urn contains 10 identical balls of which 5 are black, 3 are red, and 2 are white. Four balls are drawn one at a time and without replacement. Find the probability that the first ball is black, the second red, the third white, and the fourth black.
14 CHAPTER 2. PROBABILITY 14 Denoting by B 1 the event that the first ball is black, and likewise for R 2, W 3, and B 4, the required probability is: P(B 1 R 2 W 3 B 4 = P(B 4 B 1 R 2 W 3 P(W 3 B 1 R 2 P(R 2 B 1 P(B 1. Assuming equally likely outcomes at each step, we have: P(B 1 = 5 10,P(R 2 B 1 = 3 9,P(W 3 B 1 R 2 = 2 8,P(B 4 B 1 R 2 W 3 = 4 7. Therefore, P(B 1 R 2 W 3 B 4 = = Answer: The events {A 1,A 2,...,A n } form a partition of S, if these events are pairwise disjoint, A i A j =, i j, and their union is S, n j=1 A j = S. Then it is obvious that any event B in S may be expressed as follows, in terms of a partition of S; namely, B = n j=1 (A j B. The concept of partition is defined similarly for countably infinite many events, and the probability P(B is expressed likewise. In the sequel, by writing j = 1, 2,... and j we mean to include both cases, finitely many indices and countably infinite many indices. Thus, we have the following result. Theorem 2.3 Total Probability Theorem Let {A 1,A 2,...} be a partition of S, and let P(A j > 0 for all j. Then, for any event B, P(B = j P(B A j P(A j. The significance of this result is that, if it happens that we know the probabilities of the partitioning events P(A j, as well as the conditional probabilities of B, given A j, then these quantities may be combined, according to the preceding formula, to produce the probability P(B.
15 CHAPTER 2. PROBABILITY 15 Example 2.7 The proportion of motorists in a given gas station using regular unleaded gasoline, extra unleaded, and premium unleaded over a specified period of time are 40%, 35%, and 25%, respectively. The respective proportions of filling their tanks are 30%, 50%, and 60%. What is the probability that a motorist selected at random from among the patrons of the gas station under consideration and for the specified period of time will fill his/her tank? Denote by R, E, and P the events of a motorist using unleaded gasoline which is regular, extra unleaded, and premium, respectively, and by F the event of having the tank filled. Then the translation into terms of probabilities of the proportions given above is: P(R = 0.40,P(E = 0.35,P(P = 0.25, P(F R = 0.30,P(F E = 0.50,P(F P = Then the required probability is: P(F = P(F RP(R + P(F EP(E + P(F PP(P = = Answer: Exercises Exercise 2.5 For the events A, B, C and their complements, suppose that: P(A B C = 1, 16 P(A B c C = 5, P(A B 16 Cc = 3, 16 P(A Bc C c = 2, 16 P(Ac B C = 2, 16 P(A c B C c = 1, 16 P(Ac B c C = 1, and 16 P(Ac B c C c = Calculate the probabilities: P(A, P(B, P(C. 2. Determine whether or not the events A, B, and C are independent. 3. Determine whether or not the events A and B are independent. 1. A = (A B C (A B c C (A B C c (A B c C c and hence P(A = Likewise, P(B = , P(C =
16 CHAPTER 2. PROBABILITY = 1 = P(A B C P(AP(BP(C = Thus A, B, and 16 C are not independent. 3. P(A B = P(A B C+P(A B C c = = A and B are not independent. P(AP(B = , Answer: , , Bayes theorem The question arises of whether experimentation may lead to reevaluation of the prior probabilities on the basis of new evidence. To put it more formally, is it possible to use P(A j and P(B A j, j = 1, 2,... in order to calculate P(A j B? Theorem 2.4 Bayes Formula Let {A 1,A 2,...} be a partition of S, and let P(A j > 0 for all j. Then, for any j = 1, 2,...: P(A j B = P(B A jp(a j i P(B A ip(a i. Indeed, P(A j B = P(A j B completes the proof. = P(B A jp(a j P(B P(B, and then Total Probability Theorem Example 2.8 A student sits a multiplechoice exam. Consider a single question with 5 possible answers. Let C be the event that the student answers the question correctly. Suppose you think the student has not been revising properly and that there is only a 30% chance that he will know the answer. Call this event K, i.e. P(K = Note that C K because even if he does know the right answer in exam conditions he may give the wrong answer although this is unlikely and if he does not know the right answer he might answer correctly by guessing. Suppose you assess P(C K = 0.95 and P(C K = 0.20 (since there are 5 answers and if he guesses he will be correct 20% of the time. Suppose the candidate answers the question correctly, what is the probability that he knew the right answer?
17 CHAPTER 2. PROBABILITY 17 P(C = P(C KP(K + P(C K c P(K c = = P(K C = P(C KP(K P(C = = Answer: Exercises Exercise 2.6 Three machines I, II, and III manufacture 30%, 30%, and 40%, respectively, of the total output of certain items. Of them, 4%, 3%, and 2%, respectively, are defective. One item is drawn at random from the total output and is tested. 1. What is the probability that the item is defective? 2. If it is found to be defective, what is the probability the item was produced by machine I? 3. Same question as in part 2 for each one of the machines II and III. 1. P(D = P(IP(D I+P(IIP(D II+P(IIIP(D III = = 0.029; 2. P(I D = P(D IP(I P(D = ; 3. P(II D = P(D IIP(II = P(D P(III D = P(D IIIP(III = P(D Answer: 0.029; 0.414; 0.310; Exercise 2.7 A bag contains 3 red and 4 white balls, a second bag contains 1 red and 5 white balls. A bag is selected at random. What is the probability that a ball drawn from this bag is white? If the ball is white what is the probability that the first bag was selected?
18 CHAPTER 2. PROBABILITY 18 Let R be an event that we draw a red ball, W that we draw a white ball, F that we select a first bag and S that we select the second bag. Then: P(R F = 3 7, P(W F = 4 7, P(R S = 1 6, P(W S = 5 6 Also we select bag at random, so each back is equally likely to be picked: P(F = 1 2, P(S = 1 2. We require P(W and P(F W P(W = P(W FP(F + P(W SP(S = = Applying Bayes theorem: P(F W = P(W FP(F P(W = 4/7 1/2 59/84 = Answer: 59/84; 24/59. Exercise 2.8 Bag A contains two white and two black balls; Bag B contains three white and two black balls. One ball is drawn from A and is transferred to B, one ball is then drawn from B and turns out to be white. What is the probability that the transferred ball was white? Let W= Ball drawn is white, T= Ball transfered is white. We know P(W T = 2 3, P(W T c = 1 2, P(T = 1 2 We require P(T W. By Bayes theorem P(T W = P(W TP(T P(W TP(T + P(W T c P(T c = 2/3 1/2 2/3 1/2 + 1/2 1/2 = 4 7 Answer: 4/7. Exercise 2.9
19 CHAPTER 2. PROBABILITY 19 A screening programme for a disease is proposed. It is thought that 1 in of the population has the disease. The screening test gives either a positive or negative result with P(+ D = 0.98 P(+ D c = 0.05 where D denotes the event the disease is present. Suppose an individual gives a positive result. What is the probability he/she has the disease, i.e. what is P(D +? Applying Bayes rule P(D + = Answer: P(+ DP(D P(+ DP(D + P(+ D c P(D c = =
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