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 Ac , is the event defined by: c A = {s ∈ S; s ∈ / A}. So Ac occurs whenever A does not, and vice versa.

S The union of the Sevents A1 , ..., An , denoted by A1 ∪ ... ∪ An or nj=1 Aj , is n the event defined Sn by j=1 Aj = {s ∈ S; s ∈ Aj , for at least one j = 1, ..., n}. So the event j=1 Aj occurs whenever at least one of Aj , j = 1, ..., n occurs. The 1

CHAPTER 2. PROBABILITY

2

definition extends to an infinite number S∞ of events. Thus, for countably infinite many events Aj , j = 1, 2, ..., one has j=1 Aj = {s ∈ S; s ∈ Aj , for at least one j = 1, 2, ...}.

The intersection of the events T Aj , j = 1, ..., n is the event denoted by A1 ∩ T ... ∩ AnT or nj=1 Aj and is defined by nj=1 Aj = {s ∈ S; s ∈ Aj , for all j = 1, ..., n}. Thus, nj=1 Aj occurs whenever all Aj , j = 1, ..., n occur simultaneously. This definition extends to an infinite number of events. Thus, for countably infinite T many events Aj , j = 1, 2, ..., one has ∞ A j=1 j = {s ∈ S; s ∈ Aj , for all j = 1, 2, ...}.

If A1 ∩ A2 = ∅, the events A1 and A2 are called disjoint. The events Aj , j = 1, 2, ..., are said to be mutually or pairwise disjoint, if Ai ∩Aj = ∅ whenever i 6= j.

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3

The differences A1 − A2 and A2 − A1 are the events defined by A1 − A2 = {s ∈ S; s ∈ A1 , s ∈ / A2 }, A2 − A1 = {s ∈ S; s ∈ A2 , s ∈ A1 }.

From the definition of the preceding operations, the following properties follow immediately 1. S c = ∅, ∅c = S, (Ac )c = A. 2. S ∪ A = S, ∅ ∪ A = A, A ∪ Ac = S, A ∪ A = A. 3. S ∩ A = A, ∅ ∩ A = ∅, A ∩ Ac = ∅, A ∩ A = A. 4. Commutative laws A1 ∪ A2 = A2 ∪ A1 A1 ∩ A2 = A2 ∩ A1 5. Associative laws A1 ∪ (A2 ∪ A3 ) = (A1 ∪ A2 ) ∪ A3 A1 ∩ (A2 ∩ A3 ) = (A1 ∩ A2 ) ∩ A3 6. An identity: ∪j Aj = A1 ∪ (Ac1 ∩ A2 ) ∪ (Ac1 ∩ Ac2 ∩ A3 ) ∪ ... 7. Distributive laws) A ∩ (∪j Aj ) = ∪j (A ∩ Aj ) A ∪ (∩j Aj ) = ∩j (A ∪ Aj ) 8. DeMorgan s laws: (∪j Aj )c = ∩j Acj (∩j Aj )c = ∪j Acj 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

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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 x1 , ..., xn , or countably in infinite many values such as x1 , x2 , .... 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) ∈ R2 ; −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). Solution 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 Ai , i = 1, 2, ..., T Ai Aj = ∅ , i 6= j, it holds ! ∞ ∞ [ X Ai = P P (Ai ) i=1

i=1

Next, we present some basic results following immediately from the defining properties of the probability: 1. For an empty set P (∅) = 0 Proof 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

S P 2. For any pairwise disjoint events A1 , ..., An , P ( ni=1 Ai ) = ni=1 P (Ai ).

Proof

Take Ai = ∅ for i ≥ n + 1, consider the following obvious relation, and use (P3) and #1 to obtain: ! ! ∞ n ∞ n X X [ [ P (A1 ) = P (Ai ). P Ai = Ai = P i=1

i=1

i=1

i=1

△ 3. For any event A, P (Ac ) = 1 − P (A). Proof From (P2) and #2, P (A ∪ Ac ) = P (S) = 1 or P (A) + P (Ac ) = 1, so that P (Ac ) = 1 − P (A). △ 4. A1 ⊆ A2 implies P (A1 ) ≤ P (A2 ) and P (A2 − A1 ) = P (A2 ) − P (A1 ). Proof The relation A1 ⊆ A2 , implies A2 = A1 ∪ (A2 − A1 ), so that, by #2, P (A2 ) = P (A1 ) + P (A2 − A1 ). Solving for P (A2 − A1 ), we obtain P (A2 − A1 ) = P (A2 ) − P (A1 ), so that, by (P1), P (A1 ) ≤ P (A2 ). At this point it must be pointed out that P (A2 − A1 ) need not be P (A2 ) − P (A1 ), if A1 is not contained in A2 . △ 5. 0 ≤ P (A) ≤ 1 for every event A. Proof Clearly, ∅ ⊆ A ⊆ S for any event A. Then (P1), #1 and #4 give: 0 = P (∅) ≤ P (A) ≤ P (S) = 1. △ 6. (a) For any two events A1 and A2 : P (A1 ∪ A2 ) = P (A1 ) + P (A2 ) − P (A1 ∩ A2 ). Proof It is clear (by means of a Venn diagram, for example) that A1 ∪ A2 = A1 ∪ (A2 ∩ Ac1 ) = A1 ∪ (A2 − A1 ∩ A2 ).

6

CHAPTER 2. PROBABILITY Then, by means of #2 and #4: P (A1 ∪ A2 ) = P (A1 ) + P (A2 − A1 ∩ A2 ) = P (A1 ) + P (A2 ) − P (A1 ∩ A2 ).

△ 6. (b) For any three events A1 , A2 , and A3 : P (A1 ∪ A2 ∪ A3 ) = P (A1 ) + P (A2 ) + P (A3 ) − (P (A1 ∩ A2 ) +P (A1 ∩ A3 ) + P (A2 ∩ A3 )) + P (A1 ∩ A2 ∩ A3 ). Proof Apply part (a) to obtain: P (A1 ∪ A2 ∪ A3 ) = P [(A1 ∪ A2 ) ∪ A3 ]

= P (A1 ∪ A2 ) + P (A3 ) − P [(A1 ∪ A2 ) ∩ A3 ]

= P (A1 ) + P (A2 ) − P (A1 ∩ A2 ) + P (A3 ) − P [(A1 ∩ A3 ) ∪ (A2 ∩ A3 )]

= P (A1 ) + P (A2 ) + P (A3 ) − P (A1 ∩ A2 )

− [P (A1 ∩ A3 ) + P (A2 ∩ A3 ) − P (A1 ∩ A2 ∩ A3 )]

= P (A1 )+P (A2 )+P (A3 )−P (A1 ∩A−2)−P (A1 ∩A3 )−P (A2 ∩A3 )+P (A1 ∩A2 ∩A3 ). △ S∞ P∞ Sn 7. For any events A , A , ..., P ( A ) ≤ P (A ), and P ( 1 2 i i i=1 i=1 i=1 Ai ) ≤ Pn i=1 P (Ai ).

Proof

By the identity given in the start of the section and (P3): ! ∞ [ 
 Ai = P A1 ∪ (Ac1 ∩ A2 ) ∪ ... ∪ Ac1 ∩ ... ∩ Acn−1 ∩ An ∪ ... P i=1

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CHAPTER 2. PROBABILITY


= P (A1 ) + P (Ac1 ∩ A2 ) + ... + P Ac1 ∩ ... ∩ Acn−1 ∩ An + ... ≤ P (A1 ) + P (A2 ) + ... + P (An ) + ... (by #4). For the finite case:

P

n [

Ai

i=1

!


  = P A1 ∪ (Ac1 ∩ A2 ) ∪ . . . ∪ Ac1 ∩ . . . ∩ Acn−1 ∩ An

= P (A1 ) + P (Ac1 ∩ A2 ) + ... + P Ac1 ∩ ... ∩ Acn−1 ∩ An




≤ P (A1 ) + P (A2 ) + ... + P (An ). △ 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, A1 , ..., An , is given by: ! n n [ X X Aj = P P (Aj ) − P (Aj1 ∩ Aj2 ) j=1

+

X

j=1

1≤j1 {HHT } = A ∩ B occurred. The required probability is then 31 = 1/8 3/8 (B) 1 = P (A). 4 △

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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 (uncondi= P (A). Thus, the conditional probability is a tional) probability of A, as P P(A∩S) (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) = = 1; P (B) P (B)

and if A1 , A2 , ... are pairwise disjoint, then ! 
   ∞ [ P ∪∞ P ∪∞ j=1 Aj ∩ B j=1 (Aj ∩ B) Aj |B = P = P (B) P (B) j=1 P∞ ∞ ∞ X P (Aj ∩ B) X j=1 P (Aj ∩ B) = = = P (Aj |B). P (B) P (B) j=1 j=1 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 P (B|A) =

P (B ∩ A) P (A ∩ B) P (A|B)P (B) P (A)P (B) = = = = P (B). P (A) P (A) P (A) P (A)

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 (A)P (B). Two events A1 and A2 are said to be independent (statistically or in the probability sense), if P (A1 ∩A2 ) = P (A1 )P (A2 ). When P (A1 ∩A2 ) 6= P (A1)P (A2) they are said to be dependent. The definition of independence can be extended to any number n of events A1 , ..., An by requiring any combination of the events to be independent. The events A1 , ..., An 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 j1 , ..., jk with 1 ≤ j1 < < jk ≤ n, we have: ! k k \ Y Aji = P P (Aji ). i=1

Example 2.5

i=1

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13

Let S = {1, 2, 3, 4} and let P ({1}) = P ({2}) = P ({3}) = P ({4}) = 1/4. Define the events A1 , A2 , A3 by: A1 = {1, 2}, A2 = {1, 3}, A3 = {1, 4}. Check if events A1 , A2 and A3 are independent. Solution P (A1 ) = P (A2 ) = P (A3 ) = 12 P (A1 ∩ A2 ) = P ({1}) = 14 = P (A1 ) × P (A2 ), P (A1 ∩ A3 ) = P ({1}) = 14 = P (A1 ) × P (A3 ), P (A2 ∩ A3 ) = P ({1}) = 14 = P (A2 ) × P (A3 ). However, P (A1 ∩ A2 ∩ A3 ) = P ({1}) = 14 6= P (A1 )P (A2 )P (A3 ) = 18 . 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 Tn−1 For any n events A1 , ..., An with P ( j=1 Aj ) > 0, it holds: P

n \

Aj

j=1

!

= P (An |A1 ∩ ... ∩ An−1 )P (An−1 |A1 ∩ ... ∩ An−2 )

...P (A2 |A1 )P (A1 ). Proof 1 ∩P2 ) yields P (A1 ∩ A2 ) = For n = 2, the theorem is true since p(A2 |A1 ) = P (A P (A1 ) P (A2 |A1 )P (A1 ). Next, assume P (A1 ∩ ... ∩ Ak ) = P (Ak |A1 ∩ ... ∩ Ak )...P (A2 |A1 )P (A1 ) and show that P (A1 ∩ ...Ak+1 ) = P (Ak+1 |A1 ∩ ... ∩ Ak )P (Ak |A1 ∩ ... ∩ Ak−1 )...P (A2 |A1 )p(A1 ). Indeed, P (A1 ∩...∩Ak+1 ) = P ((A1 ∩...∩Ak )∩Ak+1 ) = P (Ak+1 |A1 ∩...∩Ak )P (A1 ∩ ... ∩ Ak ) = P (Ak+1 |A1 ∩ ... ∩ Ak )P (Ak |A1 ∩ ... ∩ Ak−1 )...P (A2 |A1 )P (A1 ) 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. Solution

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14

Denoting by B1 the event that the first ball is black, and likewise for R2 , W3 , and B4 , the required probability is: P (B1 ∩ R2 ∩ W3 ∩ B4 ) = P (B4 |B1 ∩ R2 ∩ W3 )P (W3 |B1 ∩ R2 )P (R2 |B1 )P (B1 ). Assuming equally likely outcomes at each step, we have: 5 , P (R2 |B1 ) = 39 , P (W3 |B1 ∩R2 ) = 82 , P (B4 |B1 ∩R2 ∩W3 ) = 74 . Therefore, P (B1 ) = 10 5 1 P (B1 ∩ R2 ∩ W3 ∩ B4 ) = 47 × 82 × 39 × 10 = 42 ≈ 0.024. △ Answer: 0.024 The events {A1 , A2 , ..., An } form a partition of S S, if these events are pairwise disjoint, Ai ∩ Aj = ∅, i 6= j, and their union is S, nj=1 Aj = S. Then it is obvious that any event S B in S may be expressed as follows, in terms of a partition of S; namely, B = nj=1 (Aj ∩ B).

The concept of partition is defined similarly for countably infinite many events, and the P 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 {A1 , A2 , ...} be a partition of S, and let P (Aj ) > 0 for all j. Then, for any event B, X P (B) = P (B|Aj )P (Aj ). j

The significance of this result is that, if it happens that we know the probabilities of the partitioning events P (Aj ), as well as the conditional probabilities of B, given Aj , then these quantities may be combined, according to the preceding formula, to produce the probability P (B).

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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? Solution 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 ) = 0.60. Then the required probability is: P (F ) = P (F |R)P (R) + P (F |E)P (E) + P (F |P )P (P ) = 0.30 × 0.40 + 0.50 × 0.35 + 0.60 × 0.25 = 0.445. △ Answer: 0.4455.

Exercises Exercise 2.5 For the events A, B, C and their complements, suppose that: P (A ∩ B ∩ C) = 3 2 5 , P (A ∩ B ∩ C c ) = 16 , P (A ∩ B c ∩ C c ) = 16 , P (Ac ∩ B ∩ C) = P (A ∩ B c ∩ C) = 16 1 1 1 c c c c c c P (A ∩ B ∩ C ) = 16 , P (A ∩ B ∩ C) = 16 , and P (A ∩ B ∩ C c ) = 16 .

1 , 16 2 , 16

1. 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. Solution 1. A = (A ∩ B ∩ C) ∪ (A ∩ B c ∪ C) ∪ (A ∩ B ∩ C c ) ∪ (A ∩ B c ∩ C c ) and hence P (A) = 0.6875. Likewise, P (B) = 0.4375, P (C) = 0.5625.

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CHAPTER 2. PROBABILITY

1 = P (A ∩ B ∩ C) 6= P (A)P (B)P (C) = 0.1691. Thus A, B, and 2. 0.0625 = 16 C are not independent.

3. P (A∩B) = P (A∩B∩C)+P (A∩B∩C c ) = A and B are not independent.

1 +3 16 16

=

1 4

6= P (A)P (B) = 0.3001, △

Answer: 0.6875, 0.4375, 0.5625.

2.3

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 (Aj ) and P (B|Aj ), j = 1, 2, ... in order to calculate P (Aj |B)? Theorem 2.4 Bayes Formula Let {A1 , A2 , ...} be a partition of S, and let P (Aj ) > 0 for all j. Then, for any j = 1, 2, ...: P (B|Aj )P (Aj ) P (Aj |B) = P . i P (B|Ai )P (Ai ) Proof Indeed, P (Aj |B) = completes the proof.

P (Aj ∩B) P (B)

=

P (B|Aj )P (Aj ) , P (B)

and then Total Probability Theorem △

Example 2.8 A student sits a multiple-choice 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) = 0.30. Note that C 6= 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 ¯ = 0.20 (since there are 5 answers and if assess P (C | K) = 0.95 and P (C | K) 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? Solution

17

CHAPTER 2. PROBABILITY

P (C) = P (C|K)P (K) + P (C|K c )P (K c ) = 0.95 × 0.3 + 0.2 × 0.7 = 0.425 P (C|K)P (K) 0.95 × 0.3 = = 0.671 P (C) 0.425

P (K|C) =

△ Answer: 0.671

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. Solution 1. P (D) = P (I)P (D|I)+P (II)P (D|II)+P (III)P (D|III) = 0.3×0.04+0.3× 0.03 + 0.4 × 0.02 = 0.029; 2. P (I|D) =

P (D|I)P (I) P (D)

=

0.040.3 0.029

P (D|II)P (II) = 0.03×0.3 ≈ P (D) 0.029 P (D|III)P (III) 0.02×0.4 = 0.029 = P (D)

3. P (II|D) = P (III|D)

≈ 0.414; 0.310 ≈ 0.276 △

Answer: 0.029; 0.414; 0.310; 0.276 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 Solution

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: 3 P (R|F ) = , 7

4 P (W |F ) = , 7

1 P (R|S) = , 6

P (W |S) =

5 6

Also we select bag at random, so each back is equally likely to be picked: 1 P (S) = . 2

1 P (F ) = , 2

We require P (W ) and P (F |W ) P (W ) = P (W |F )P (F ) + P (W |S)P (S) =

4 1 5× 59 × + 2= 7 2 61 84

Applying Bayes’ theorem: P (F |W ) =

4/7 × 1/2 24 P (W |F )P (F ) = = P (W ) 59/84 59 △

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? Solution Let W = ’Ball drawn is white’, T = ’Ball transfered is white’. We know 2 P (W |T ) = , 3

1 P (W |T c ) = , 2

P (T ) =

1 2

We require P (T |W ). By Bayes’ theorem P (T |W ) =

P (W |T )P (T ) 2/3 × 1/2 4 = = c c P (W |T )P (T ) + P (W |T )P (T ) 2/3 × 1/2 + 1/2 × 1/2 7 △

Answer: 4/7. Exercise 2.9

CHAPTER 2. PROBABILITY

19

A screening programme for a disease is proposed. It is thought that 1 in 10000 of the population has the disease. The screening test gives either a positive or negative result with P (+|D) = 0.98

P (+|Dc ) = 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|+)? Solution Applying Bayes’ rule P (D|+) =

P (+|D)P (D) 0.98 × 0.0001 = = 1.96×10−3 c c P (+|D)P (D) + P (+|D )P (D ) 0.98 × 0.0001 + 0.05 × 0.9999 △

Answer: 1.96 × 10−3