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}