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.
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 3 2 3 = 1 P(B C) ; P(B C) = = 2 P(C) 1 6 2 3 = 1 4.
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 3 2 3 = 1 P(B C) ; P(B C) = = 2 P(C) 1 6 2 3 = 1 4. P(A C) = P(A) while P(B C) P(B)
Definition Two events A and B are independent if P(A B) = P(A), and are dependent otherwise.
Definition Two events A and B are independent if P(A B) = P(A), and are dependent otherwise. Remark:
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.
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)
Remark:
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.
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.
The Multiplication Rule for Independent Events
The Multiplication Rule for Independent Events The general multiplication rule tells us P(A B) = P(A B) P(B).
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).
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)
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.
In real life, we often use this multiplication rule without noticing it.
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 1 2 1 2 ;
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 1 2 1 2 ; 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 1 6 1 6 1 6 1 6 1 6 1 6 ;
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 1 2 1 2 ; 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 1 6 1 6 1 6 1 6 1 6 1 6 ; 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 1 4 1 4 1 4.
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 1 2 1 2 ; 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 1 6 1 6 1 6 1 6 1 6 1 6 ; 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 1 4 1 4 1 4. 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 }) = 12 51.
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.
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.
Remark:
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)
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.
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 ).
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.
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?
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?
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.