This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: AMS 410 Actuarial Mathematics Fall 2009 Exercise 2: General Probability II 09/10/2009 1. A test for a disease correctly diagnoses a diseased person as having the disease with probability 85%. The test incorrectly diagnoses someone without the disease as having the disease with a probability of 10%. If 1% of the people in a population have the disease: (a) What's the chance that a person from this population tests positive for the disease? (Law of Total Probability) (b) What's the chance that a person from this population who tests positive for the disease actually has the disease? (Bayes' Rule) 2. Let A , B ,and C be events such that A and B are independent, B and C are mutually exclu- sive, P [ A ] = 1 4 , P [ B ] = 1 6 , and P [ C ] = 1 2 . Calculate P [( A ∩ B ) ∪ C ] . (Independence) 1 3. The following probabilities of three events in a sample space are given: P [ A ] =0.6, P [ B ] =0.5, P [ C ] =0.4; P [ A ∪ B ] =1, P [ A ∪ C ] =0.7, P [ B ∪ C ] =0.7. Calculate P [ A ∩...
View Full Document
- Fall '08
- Conditional Probability, Probability theory, Orthostatic hypotension