University of New Hampshire Experiments and Events Probability Questions

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Mathematics

University of New Hampshire

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• Unless otherwise specified, you can leave your answer in closed form (e.g. 1 – (120)(0.1)200). • Show your work. Answers without justification will be given little credit. Justify each step in your solutions e.g. by stating that the step follows from an axiom of probability, a definition, algebra, etc.; for example, your answer could include a line like this: Pr(XnYn Z) · Pr(AUB) = Pr(XnYN Z) · (Pr(A) + Pr(B)) (A and B are disjoint) PROBLEM 4. Consider three events E1, E2, E3. Recall from class that we can use set notation and operations to express combinations of events. Express each of the following events as the appropriate combination of E1, E2, E3: (a) Ej and Ez occur, but not E2. (b) None of the events occur. (c) At least one of the events occurs. (d) All of the events occur. (e) Either E1 or E2 occurs, but not both of them. (f) Only one of the events occurs. Solution: Your solution here. PROBLEM 5 (Fair die). Suppose we roll a fair die n times. Each outcome is equally likely. Find the probability of getting: (a) All sixes. (b) No sixes. (c) Not all sixes. (d) At least one six. Solution: Your solution here. PROBLEM 6. Merlin has a special coin that comes up heads with probability p and it comes up tails with probability q. (a) What conditions should p and q meet in order for the probability function to be valid? (b) Merlin has determined that, when tossing his special coin, heads comes up three times more often than tails. Given this information, what are the probabilities p and q for Merlin's special coin? Solution: Your solution here.
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3 Questions
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Please have a look at the solution.

4-a) (E1 ∩ E3) – E2
4-b) U – (E1 ∪ E2 ∪ E3) (U is the universal set)
4-c) E1 ∪ E2 ∪ E3
4-d) E1 ∩ E2 ∩ E3
4-e) (E1 ∪ E2) – (E1 ∩ E2)
4-f) (E1 ∪ E2 ∪ E3) - [ ((E1 ∩ E2) ∪ (E1 ∩ E3) ∪ (E2 ∩ E3)) – 3 (E1 ∩ E2 ∩ E3)]
5-a) We have,
Pr(A and B) = Pr(A)*Pr(B) where A and B are independent events
Pr(getting 6 in one roll) = 1/6
So, Pr(getting sixes in all n rolls) = 1/6*1/6*1/6---n times = [1/6]n
5-b) We have,
Pr(A and B) = Pr(A)*P...


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