Determining Probabilities. Product Rule for Ordered Pairs/k-Tuples:

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1 Determining Probabilities Product Rule for Ordered Pairs/k-Tuples:

2 Determining Probabilities Product Rule for Ordered Pairs/k-Tuples: Proposition If the first element of object of an ordered pair can be selected in n 1 ways, and for each of these n 1 ways the second element of the pair can be selected in n 2 ways, then the number of pairs is n 1 n 2.

3 Determining Probabilities Product Rule for Ordered Pairs/k-Tuples: Proposition If the first element of object of an ordered pair can be selected in n 1 ways, and for each of these n 1 ways the second element of the pair can be selected in n 2 ways, then the number of pairs is n 1 n 2. Proposition Suppose a set consists of ordered collections of k elements (k-tuples) and that there are n 1 possible choices for the first element; for each choice of the first element, there n 2 possible choices of the second element;... ; for each possible choice of the first k 1 elements, there are n k choices of the k th element. Then there are n 1 n 2 n k possible k-tuples.

4 Determining Probabilities Proposition P k:n = n (n 1) (n (k 1)) = where k! = k (k 1) 2 1 is the k factorial. n! (n k)!

5 Determining Probabilities Proposition P k:n = n (n 1) (n (k 1)) = where k! = k (k 1) 2 1 is the k factorial. Proposition ( ) n = P k:n k k! = n! k!(n k)! where k! = k (k 1) 2 1 is the k factorial. n! (n k)!

6 Conditional Probability

7 Conditional Probability Definition For any two events A and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by P(A B) = P(A B) P(B)

8 Conditional Probability Definition For any two events A and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by P(A B) = P(A B) P(B) Event B is the prior knowledge. Due to the presence of event B, the probability for event A to happen changed.

9 Conditional Probability Definition For any two events A and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by P(A B) = P(A B) P(B) Event B is the prior knowledge. Due to the presence of event B, the probability for event A to happen changed. The Multiplication Rule P(A B) = P(A B) P(B)

10 Conditional Probability

11 Conditional Probability Example 2.29 A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1, 30% are brand 2, and 20% are brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of brand 1 s DVD players require warranty on parts and labor, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively. 1. What is the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty? 2. What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty?

12 Conditional Probability Example 2.29 A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1, 30% are brand 2, and 20% are brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of brand 1 s DVD players require warranty on parts and labor, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively. 1. What is the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty? 2. What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty? 3. If a customer returns to the store with a DVD player that needs warranty work, what is the probability that it is a brand 1 DVD player? A brand 2 DVD player? A brand 3 DVD player?

13 Conditional Probability

14 Conditional Probability The Law of Total Probability Let A 1, A 2,..., A k be mutually exclusive and exhaustive events. Then for any other event B, P(B) = P(B A 1 ) P(A 1 ) + P(B A 2 ) P(A 2 ) + + P(B A k ) P(A k ) k = P(B A i ) P(A i ) i=1 where exhaustive means A 1 A 2 A k = S.

15 Conditional Probability The Law of Total Probability Let A 1, A 2,..., A k be mutually exclusive and exhaustive events. Then for any other event B, P(B) = P(B A 1 ) P(A 1 ) + P(B A 2 ) P(A 2 ) + + P(B A k ) P(A k ) k = P(B A i ) P(A i ) i=1 where exhaustive means A 1 A 2 A k = S.

16 Conditional Probability

17 Conditional Probability Bayes Theorem Let A 1, A 2,..., A k be a collection of k mutually exclusive and exhaustive events with prior probabilities P(A i )(i = 1, 2,..., k). Then for any other event B with P(B) > 0, the posterior probability of A j given that B has occurred is P(A j B) = P(A j B) P(B) = P(B A j ) P(A j ) k i=1 P(B A i) P(A i ) j = 1, 2,... k

18 Conditional Probability

19 Conditional Probability Application of Bayes Theorem Example 2.30 Incidence of a rare disease Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result will occur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time. If a randomly selected individual is tested and the result is positive, what is the probability that the individual has the disease?

21 Example: A fair die is tossed and we want to guess the outcome. The outcomes will be 1, 2, 3, 4, 5, 6 with equal probability 1 6 each. If we are interested in getting the following results: A = {1, 3, 5}, B = {1, 2, 3} and C = {3, 4, 5, 6}, then we can calculate the probability for each event: P(A) = P(B) = 3 6 = 1 2, and P(C) = 4 6 = 2 3.

22 Example: A fair die is tossed and we want to guess the outcome. The outcomes will be 1, 2, 3, 4, 5, 6 with equal probability 1 6 each. If we are interested in getting the following results: A = {1, 3, 5}, B = {1, 2, 3} and C = {3, 4, 5, 6}, then we can calculate the probability for each event: P(A) = P(B) = 3 6 = 1 2, and P(C) = 4 6 = 2 3. If someone tell you that after one toss, event C happened, i.e. the outcome is one of {3, 4, 5, 6}, then what is the probability for event A to happen and what for B? P(A C) = P(A C) P(C) = = 1 P(B C) ; P(B C) = = 2 P(C) = 1 4.

23 Example: A fair die is tossed and we want to guess the outcome. The outcomes will be 1, 2, 3, 4, 5, 6 with equal probability 1 6 each. If we are interested in getting the following results: A = {1, 3, 5}, B = {1, 2, 3} and C = {3, 4, 5, 6}, then we can calculate the probability for each event: P(A) = P(B) = 3 6 = 1 2, and P(C) = 4 6 = 2 3. If someone tell you that after one toss, event C happened, i.e. the outcome is one of {3, 4, 5, 6}, then what is the probability for event A to happen and what for B? P(A C) = P(A C) P(C) = = 1 P(B C) ; P(B C) = = 2 P(C) = 1 4. P(A C) = P(A) while P(B C) P(B)

25 Definition Two events A and B are independent if P(A B) = P(A), and are dependent otherwise.

26 Definition Two events A and B are independent if P(A B) = P(A), and are dependent otherwise. Remark:

27 Definition Two events A and B are independent if P(A B) = P(A), and are dependent otherwise. Remark: 1. P(A B) = P(A) P(B A) = P(B). This is natural since the definition for independent should be symmetric.

28 Definition Two events A and B are independent if P(A B) = P(A), and are dependent otherwise. Remark: 1. P(A B) = P(A) P(B A) = P(B). This is natural since the definition for independent should be symmetric. P(B A) = P(A B) P(A) = P(A B) P(B) P(A)

30 Remark:

31 Remark: 2. If events A and B are mutually disjoint, then they can not be independent. Intuitively, if we know event A happens, we then know that B does not happen, since A B =. Mathmatically, P(A B) = P(A B) P(B) = P( ) P(B) = 0 P(A), unless P(A) = 0 which is trivial.

32 Remark: 2. If events A and B are mutually disjoint, then they can not be independent. Intuitively, if we know event A happens, we then know that B does not happen, since A B =. Mathmatically, P(A B) = P(A B) P(B) = P( ) P(B) = 0 P(A), unless P(A) = 0 which is trivial. e.g. for the die tossing example, if A = {1, 3, 5} and B = {2, 4, 6}, then P(A B) = P( ) = 0, therefore P(A B) = 0. However, P(A) = 0.5.

34 The Multiplication Rule for Independent Events

35 The Multiplication Rule for Independent Events The general multiplication rule tells us P(A B) = P(A B) P(B).

36 The Multiplication Rule for Independent Events The general multiplication rule tells us P(A B) = P(A B) P(B). However, if A and B are independent, then the above equation would be P(A B) = P(A) P(B) since P(A B) = P(A).

37 The Multiplication Rule for Independent Events The general multiplication rule tells us P(A B) = P(A B) P(B). However, if A and B are independent, then the above equation would be P(A B) = P(A) P(B) since P(A B) = P(A). Furthermore, we have the following Proposition Events A and B are independent if and only if P(A B) = P(A) P(B)

38 The Multiplication Rule for Independent Events The general multiplication rule tells us P(A B) = P(A B) P(B). However, if A and B are independent, then the above equation would be P(A B) = P(A) P(B) since P(A B) = P(A). Furthermore, we have the following Proposition Events A and B are independent if and only if P(A B) = P(A) P(B) In words, events A and B are independent iff (if and only if) the probability that the both occur (A B) is the product of the two individual probabilities.

40 In real life, we often use this multiplication rule without noticing it.

41 In real life, we often use this multiplication rule without noticing it. The probability for getting {HH} when you toss a fair coin twice is 1 4, which is obtained by ;

42 In real life, we often use this multiplication rule without noticing it. The probability for getting {HH} when you toss a fair coin twice is 1 4, which is obtained by ; The probability for getting {6,5,4,3,2,1} when you toss a fair die six times is ( 1 6 )6, which is simply obtained by ;

43 In real life, we often use this multiplication rule without noticing it. The probability for getting {HH} when you toss a fair coin twice is 1 4, which is obtained by ; The probability for getting {6,5,4,3,2,1} when you toss a fair die six times is ( 1 6 )6, which is simply obtained by ; The probability for getting { } when you draw three cards from a deck of well-shuffled cards with replacement is 1 64, which is simply obtained by

44 In real life, we often use this multiplication rule without noticing it. The probability for getting {HH} when you toss a fair coin twice is 1 4, which is obtained by ; The probability for getting {6,5,4,3,2,1} when you toss a fair die six times is ( 1 6 )6, which is simply obtained by ; The probability for getting { } when you draw three cards from a deck of well-shuffled cards with replacement is 1 64, which is simply obtained by However, if you draw the cards without replacement, then the multiplication rule for independent events fails since the event {the first card is } is no longer independent of the event {the second card is }. In fact, P({the second card is the first card is }) =

46 Example: Exercise 89 Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events C 1 ={left ear tag is lost} and C 2 = {right ear tag is lost}. Let π = P(C 1 ) = P(C 2 ), and assume C 1 and C 2 are independent events. Derive an expression (involving π) for the probability that exactly one tag is lost given that at most one is lost.

47 Example: Exercise 89 Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events C 1 ={left ear tag is lost} and C 2 = {right ear tag is lost}. Let π = P(C 1 ) = P(C 2 ), and assume C 1 and C 2 are independent events. Derive an expression (involving π) for the probability that exactly one tag is lost given that at most one is lost.

49 Remark:

50 Remark: 1. If events A and B are independent, then so are events A and B, events A and B as well as events A and B. P(A B) = P(A B) P(B) = P(B) P(A B) P(B) = 1 P(A B) = 1 P(A) = P(A ) = 1 P(A B) P(B)

51 Remark: 1. If events A and B are independent, then so are events A and B, events A and B as well as events A and B. P(A B) = P(A B) P(B) = P(B) P(A B) P(B) = 1 P(A B) = 1 P(A) = P(A ) = 1 P(A B) P(B) 2. We can use the condition P(A B) = P(A) P(B) to define the independence of the two events A and B.

53 Independence of More Than Two Events Definition Events A 1, A 2,..., A n are mutually independent if for every k (k = 2, 3,..., n) and every subset of indices i 1, i 2,..., i k, P(A i1 A i2 A ik ) = P(A ii ) P(A i2 ) P(A ik ).

54 Independence of More Than Two Events Definition Events A 1, A 2,..., A n are mutually independent if for every k (k = 2, 3,..., n) and every subset of indices i 1, i 2,..., i k, P(A i1 A i2 A ik ) = P(A ii ) P(A i2 ) P(A ik ). In words, n events are mutually independent if the probability of the intersection of any subset of the n events is equal to the product of the individual probabilities.

56 An very interesting example: Exercise 113 A box contains the following four slips of paper, each having exactly the same dimensions: (1) win prize 1; (2) win prize 2; (3) win prize 3; and (4) win prize 1, 2 and 3. One slip will be randomly selected. Let A 1 = {win prize 1}, A 2 = {win prize 2}, and A 3 = {win prize 3}. Are these three events mutually independent?

58 Example: Consider a system of seven identical components connected as following. For the system to work properly, the current must be able to go through the system from the left end to the right end. If components work independently of one another and P(component works)=0.9, then what is the probability for the system to work?

59 Example: Consider a system of seven identical components connected as following. For the system to work properly, the current must be able to go through the system from the left end to the right end. If components work independently of one another and P(component works)=0.9, then what is the probability for the system to work? Let A = {the system works} and A i = {component i works}. Then A = (A 1 A 2 ) ((A 3 A 4 ) (A 5 A 6 )) A 7.

Independence. P(A) = P(B) = 3 6 = 1 2, and P(C) = 4 6 = 2 3.

Independence. P(A) = P(B) = 3 6 = 1 2, and P(C) = 4 6 = 2 3. Example: A fair die is tossed and we want to guess the outcome. The outcomes will be 1, 2, 3, 4, 5, 6 with equal probability 1 6 each. If we are interested in getting the following results: A = {1, 3,