A probability experiment is conducted in which the sample space of the experiment is S={ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}, event F={5, 6, 7, 8, 9}, and event G={9, 10, 11, 12}. Assume that each outcome is equally likely. List the outcomes in F or G. Find P(F or G) by counting the number of outcomes in F or G. Determine P(F or G) using the general addition rule.List the outcomes in F or G. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.A. F or G = { _____ }(Use a comma to separate answers as needed.)B. F or G = { }Find P(F or G) by counting the number of outcomes in F or G. P(F or G) =________(Type an integer or a decimal rounded to three decimal places as needed.)Determine P(F or G) using the general addition rule. Select the correct choice below and fill in any answer boxes within your choice.(Type the terms of your expression in the same order as they appear in the original expression. Round to three decimal places as needed.)A. P(F or G)=_______+_______−________=_______B. P(F or G)=_______+_______=________
Accepted Solution
A:
Answer:A) F or G = {5, 6, 7, 8, 9, 10, 11, 12}; B) P(F or G) = 8/12 = 0.667Step-by-step explanation:The outcomes in event F are 5, 6, 7, 8 and 9. The outcomes in event G are 9, 10, 11 and 12. This means that the event F or G, containing all elements of both events, will be{5, 6, 7, 8, 9, 10, 11, 12}. Note that we do not count 9 twice; it only appears in the sample space once. There are 8 elements out of 12 total in the sample space; this means the probability is 8/12 = 0.667.Using the addition rule, we begin with P(F). There are 5 elements in this event; this makes P(F) = 5/12. For P(G), there are 4 elements in G; this makes P(G) = 4/12.To add them, we add P(F) and P(G) but then must take out the element counted twice; this gives us5/12+4/12-1/12 = 9/12-1/12 = 8/12 = 0.667.