Quiz #2

Answers are given in the answer boxes below. Scroll down for explantions for the answers (given in red).

Be sure to put your name on this quiz. Write the correct answers in the answer boxes below. (That is, write the requested probabilities in the column of the table labeled probability.)

Consider the act of randomly choosing a dog from the collection of 11 dogs listed below. (It is to be assumed that each of the 11 dogs has an equal chance of being chosen.)
  1. Beau is a healthy male
  2. Buster is a healthy male
  3. Grace is a healthy female
  4. Tinkerbell is a healthy female
  5. Brodie is a sick male
  6. Fido is a sick male
  7. Sam is a sick male
  8. Wesley is a sick male
  9. Amber is a sick female
  10. Carmen is a sick female
  11. Spooner is a sick female
Letting A denote the event that a healthy dog is selected and B denote the event that a female dog is selected, give the probability for each of the 2 events indicated in the table below. (Note: BC denotes the complement of B.)

event probability
A ∪ B 7/11
BC ∩ A 2/11

Since A ∪ B is the event of selecting a healthy dog or a female dog (or both --- that is a healthy female), and since 7 of the 11 dogs are either healthy or female (or both), we have that P(A ∪ B) = 7/11, a value which could also be obtained using the fact that P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 4/11 + 5/11 - 2/11. (Because 4 of the 11 dogs are healthy, and each of the 11 dogs is equally likely to be chosen, the probability of choosing a healthy dog, P(A), is just 4/11. Similarly, because 5 of the 11 dogs are female, we have that P(B) = 5/11. Since A ∩ B is the event of selecting a healthy female, and 2 of the 11 dogs are healthy females, we have that P(A ∩ B) = 2/11.)

Since BC is the event that a male dog is selected, BC ∩ A is the event that a healthy male is selected. Since 2 of the 11 dogs are healthy males, the desired probability is clearly 2/11.

Note: Some of you made the mistake of assuming independent events. You were not given that A and B are independent, and in fact they are not independent. So P(A ∩ B) ≠ P(A) P(B), and P(BC ∩ A) ≠ P(BC) P(A).