Conditional Probability in Two-Way Tables
Task 1
Putting It All Together
Task 2
Place 6 spotted and striped balls in the jar along with 5 spotted balls that are not striped. Calculate P(A|B), the probability that a randomly selected ball is spotted, given that you know it is striped. Can you explain the results?
Task 3
Place 3 spotted balls that are not striped in the container with 5 balls that are neither spotted nor striped. Why is P(A|B) undefined in this case? In general, what condition must exist if P(A|B) is undefined? Can P(B|A) be undefined? If so, under what conditions?
Task 4
Fill the container with a random assortment of balls, including at least one ball of each type, and evaluate P(A|B), the probability that a randomly selected ball is spotted, given that you know it is striped. Can you explain why P(A|B) is equal to P(A and B)/P(B)? Will this relationship always hold? If not, what are the exceptions?
Task 5
Under what conditions is P(A|B) equal to P(B|A)?