## Quiz #1

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 8 dogs listed below. (It is to be assumed that each of the 8 dogs has an equal chance of being chosen.)
1. Beau is a healthy male
2. Buster is a healthy male
3. Tinkerbell is a healthy female
4. Fido is a sick male
5. Sam is a sick male
6. Wesley is a sick male
7. Amber is a sick female
8. 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 first 5 events indicated in the table below. (Note: BC denotes the complement of B.)

event probability
A 3/8
B 3/8
BC 5/8
AB 1/8
AB 5/8
AC 5/8

(Hints: The correct answers belong to the set { -1/2, 0, 1/8, 1/5, 3/8, 1/2, 3/5, 5/8, 4/5, 7/8, 1, 3/2 }. Also, recall that for any two events, A and B, we have that P(AB) = P(A) + P(B) - P(AB), and so this rule certainly holds for the two events under consideration. (Using this allows you to check that your answers are compatible.))

Because 3 of the 8 dogs are healthy, and each of the 8 dogs is equally likely to be chosen, the probability of choosing a healthy dog, P(A), is just 3/8. Similarly, because 3 of the 8 dogs are female, we have that P(B) = 3/8. Since P(BC) = 1 - P(B), it follows that P(BC) = 1 - 3/8 = 5/8 (a value which can also be obtained directly by noting that BC is the event of not selecting a female dog, and so is the event of selecting a male dog, and noting that 5 of the 8 dogs are male). Since AB is the event of selecting a healthy female, and 1 of the 8 dogs is a healty female, we have that P(AB) = 1/8. Since AB is the event of selecting a healthy dog or a female dog (or both --- that is a healthy female), and since 5 of the 8 dogs are either healthy or female (or both), we have that P(AB) = 5/8 (a value which could also be obtained using the fact that P(AB) = P(A) + P(B) - P(AB), and making use of the probabilities previously obtained).