1a. Briefly explain why P(AB)P(B)is equivalent toP(A)P(B)forindependentevents.

1b. Say we roll two six-sided dice and count their sum. Event A is the set of outcomes: {2, 3, 4, 5, 6}, event B is the set of outcomes {6, 7, 8}, and event C is the set of outcomes {11, 12}. What is the complement of A?

a. {9, 10, 11, 12}

b. {6}

c. {7, 8, 9, 10, 11, 12}

d. {7, 8}

1c. Are the following events disjoint, independent, or neither?

Two randomly selected students have birthdays in April

1d. Based on the tree diagram, which of these could be the numerator if we were applying Bayes' Theorem to calculate P(Disease Positive | Test Positive)?

a. P(Test Positive)

b. P(Test Positive | Disease Positive) x P(Disease Positive)

c. P(Disease Positive) + P(Test Positive)

d. P(Disease Positive)

P(Test Positive | Disease Positive) P(Disease Positive) P(Test Negative | Disease Positive) P(Test Positive | Disease Negative) P(Disease Negative) P(Test Negative | Disease Negative)