diff --git a/source/main.ptx b/source/main.ptx index 8a04378..5081079 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -25,6 +25,13 @@ + + Quizzes + + + + + Recitations diff --git a/source/quizzes/quiz-01.ptx b/source/quizzes/quiz-01.ptx new file mode 100644 index 0000000..ef5f116 --- /dev/null +++ b/source/quizzes/quiz-01.ptx @@ -0,0 +1,160 @@ + + + + + Quiz 1 + + + +

+ The following work should be completed individually. + Use of notes or textbooks is not allowed. + You may use a scientific calculator, not a graphing calculator or phone app. +

+ +

+ Show all work unless instructed otherwise. +

+ + + + + + + +

+ Write either True or False for each of the following statements. + No justification is required. +

+ + + + + +

+ If A and B are any sets, then |A\cap B| \leq |A| and |A\cap B| \leq |B|. +

+ + + +

+ True. +

+ + + + + + +

+ If A and B are any events, then \Pr(A \mid B) = 1 - \Pr(B \mid A). +

+ + + +

+ False. +

+ + + + + + +

+ If A and B are any events, then \Pr(A\cap B)\Pr(B) = \Pr(A \mid B). +

+ + + +

+ False. +

+ + + + + + +

+ We find a 4-sided die with faces 0, 1, 3, and 5. + An experiment consists of rolling the die two times. +

+ + + + + +

+ Write down the sample space \Omega of all possible outcomes for this experiment. +

+ + + +

+ + \Omega = \{ \amp (0, 0), (0, 1), (0, 3), (0, 5), + \amp (1, 0), (1, 1), (1, 3), (1, 5), + \amp (3, 0), (3, 1), (3, 3), (3, 5), + \amp (5, 0), (5, 1), (5, 3), (5, 5)\} + +

+ + + + + + +

+ Let A be the event that the second roll is at least twice the value of the first roll. + Let m B be the event that the sum of the rolls is odd. + Assume the die is fair. + List the outcomes in A and calculate \Pr(A). + List the outcomes in B and calculate \Pr(B). +

+ + + +

+ + A \amp = \{(0, 0), (0, 1), (0, 3), (0, 5), (1, 3), (1, 5)\} + B \amp = \{ (0, 1), (0, 3), (0, 5), (1, 0), (3, 0), (5, 0) \} + + Then \Pr(A) = \frac{|A|}{|\Omega|} = \frac{6}{16} = \frac{3}{8}, and \Pr(B) = \frac{|B|}{|\Omega|} = \frac{6}{16} = \frac{3}{8}. +

+ + + + + + + + + +

+ A diagnostic test is developed to detect a disease present in 1.3% of the population. + For a patient who has the disease, the test will accurately give a positive result 62% of the time. + When the patient does not have the disease, the test will accurately give a negative result 99.4% of the time. +

+ +

+ For a patient who receives a positive test, what is the probability they have the disease? +

+ + + +

+ Let P be the event of receiving a positive test result and D be the event of having the disease. + The given information is: \Pr(D) = 0.013, \Pr(P\mid D) = 0.62, and \Pr(P^c \mid D^c) = 0.994. + Then, using Bayes' Theorem: + + \Pr(D\mid P) = \frac{\Pr(P \mid D)\Pr(D)}{\Pr(P)} \amp = \frac{\Pr(P \mid D)\Pr(D)}{\Pr(P \mid D)\Pr(D) + \Pr(P \mid D^c)\Pr(D^c)} + \amp = \frac{\Pr(P \mid D)\Pr(D)}{\Pr(P \mid D)\Pr(D) + (1 - \Pr(P^c \mid D^c))(1 - \Pr(D))} + \amp = \frac{(0.62)(0.013)}{(0.62)(0.013) + (1 - 0.994)(1 - 0.013)} + \amp \approx \boxed{0.576} + +

+ + + + \ No newline at end of file