HOMEWORK 6, CS 480, due May 2nd, Jana Kosecka 1. 13.6. (10
... you tested positive for a serious disease, and that the test is 99 % accurate (i.e., the probability of testing positive given that you have the disease is 0.99, as is the probability of testing negative given that you don’t have the disease). The good news is that this is a rare disease, striking o ...
... you tested positive for a serious disease, and that the test is 99 % accurate (i.e., the probability of testing positive given that you have the disease is 0.99, as is the probability of testing negative given that you don’t have the disease). The good news is that this is a rare disease, striking o ...
CHAPTER 12 Calculator Notes for the TI-83 and TI
... f. If you press TRACE , you have the option to save the data into four lists: ROLL for the roll number, D1 for the numbers on die 1, D2 for the numbers on die 2, and SUM for the sum of the dice. Press GRAPH to save the data, or press Y to escape without saving. g. Exit the program by pressing Y to ...
... f. If you press TRACE , you have the option to save the data into four lists: ROLL for the roll number, D1 for the numbers on die 1, D2 for the numbers on die 2, and SUM for the sum of the dice. Press GRAPH to save the data, or press Y to escape without saving. g. Exit the program by pressing Y to ...
Inclusion exclusion principle
... because: if x is not in A, then all factors are 0 - 0 = 0; and otherwise, if x does belong to some Am, then the corresponding mth factor is 1 − 1 = 0. By expanding the product on the left-hand side, equation (*) follows. Use of (*): To prove the inclusion–exclusion principle for the cardinality of s ...
... because: if x is not in A, then all factors are 0 - 0 = 0; and otherwise, if x does belong to some Am, then the corresponding mth factor is 1 − 1 = 0. By expanding the product on the left-hand side, equation (*) follows. Use of (*): To prove the inclusion–exclusion principle for the cardinality of s ...
Independent and Dependent Events 10.2
... Finding Probabilities of Events In Example 1, it makes sense that the events are independent because the second guess should not be affected by the first guess. In Example 2, however, the selection of the second person depends on the selection of the first person because the same person cannot be se ...
... Finding Probabilities of Events In Example 1, it makes sense that the events are independent because the second guess should not be affected by the first guess. In Example 2, however, the selection of the second person depends on the selection of the first person because the same person cannot be se ...
Chapter 3: Probability - Wright State University
... occur. The union is denoted A ∪ B. The key word for union is “or.” Definition. The intersection of two events A and B is the event that both A and B occurs. The intersection is denoted by A ∩ B. The keyword indicating an intersection is “and”. Using the previous example, A ∪ B = {F M M, M F M, M M F ...
... occur. The union is denoted A ∪ B. The key word for union is “or.” Definition. The intersection of two events A and B is the event that both A and B occurs. The intersection is denoted by A ∩ B. The keyword indicating an intersection is “and”. Using the previous example, A ∪ B = {F M M, M F M, M M F ...
Sect. 1.5: Probability Distribution for Large N
... • Poisson Distribution: An approximation to the binomial distribution for the SPECIAL CASE when the average number (mean µ) of successes is very much smaller than the possible number n. i.e. µ << n because p << 1. • This distribution is important for the study of such phenomena as radioactive decay. ...
... • Poisson Distribution: An approximation to the binomial distribution for the SPECIAL CASE when the average number (mean µ) of successes is very much smaller than the possible number n. i.e. µ << n because p << 1. • This distribution is important for the study of such phenomena as radioactive decay. ...
