1. ## choosing exam questions

An exam has 6 essay questions and 4 multiple choice questions. A student must select at least one essay question and at least one multiple choice question.

a. a student last quarter tried to calculate this result by 6 * 4 * 8, that is choosing one essay question, one mc question, and choosing a bonus question from the remaining 8 ?s. This does not count all of the possibilities. Give a specific example of a result that is not counted by this method.

b. how many options does a student have for selecting the three questions and designating one of them to be the bonus question?

I have been thinking about this one for a while...anyone able to point me in the right direction?

2. Originally Posted by ihavvaquestion
An exam has 6 essay questions and 4 multiple choice questions. A student must select at least one essay question and at least one multiple choice question.

a. a student last quarter tried to calculate this result by 6 * 4 * 8, that is choosing one essay question, one mc question, and choosing a bonus question from the remaining 8 ?s. This does not count all of the possibilities. Give a specific example of a result that is not counted by this method.

b. how many options does a student have for selecting the three questions and designating one of them to be the bonus question?

I have been thinking about this one for a while...anyone able to point me in the right direction?
I am confused by the wording of this question. At the beginning of the problem, we are told that a student must choose at least one essay question and at least one MC question, but there is no other restriction. Then in 1(a) and 1(b) there is a notion of bonus question, and that a student has to choose exactly three questions?? This is a very odd.

My guess is that what the question designer meant is:

1(a) Suppose the student does the first essay question and the second essay question and the first MC question and then designates the MC question as bonus. This is not counted in 6*4*8.

1(b) We'd just use the original 6*4*8 and then add on the possibility that there is a sole MC bonus question, and that there is a sole essay bonus question, as such:

6*4*8 + C(6,2)*4 + C(4,2)*6

3. oops, i left out part of the question. The question should have read: A student must select 3 questions to answer and must designate one of those three questions as a bonus question.

So I understand the 6 * 4 * 8 part, but I still do not understand how this does not count all possibilities

4. Originally Posted by ihavvaquestion
oops, i left out part of the question. The question should have read: A student must select 3 questions to answer and must designate one of those three questions as a bonus question.

So I understand the 6 * 4 * 8 part, but I still do not understand how this does not count all possibilities
What if the bonus question is one of the first two that you picked?

5. ok, so if the bonus question comes from one of the first two questions picked, then would the total number of combinations would be: 6*4*8 + (6C2 * 4) + (4C2 *6)???

6. Originally Posted by ihavvaquestion
ok, so if the bonus question comes from one of the first two questions picked, then would the total number of combinations would be: 6*4*8 + (6C2 * 4) + (4C2 *6)???
I see that I should have explained my notation.

C(n, k) is another way to write nCk.

7. i understood the notation...i think...was my reply not correct?

6*4*8 + (4) 6C2 + (6) 4C2

8. Originally Posted by undefined
1(b) We'd just use the original 6*4*8 and then add on the possibility that there is a sole MC bonus question, and that there is a sole essay bonus question, as such:

6*4*8 + C(6,2)*4 + C(4,2)*6
Originally Posted by ihavvaquestion
ok, so if the bonus question comes from one of the first two questions picked, then would the total number of combinations would be: 6*4*8 + (6C2 * 4) + (4C2 *6)???

9. hey undefined...i've been thinking about this problem, and doesnt the fundamental counting principle say that we should multiply (4)(6)(8) * 6C2*4 * 4C2*6 instead of adding them on?

why do we add the last two on and not multiply if we are just counting?

10. Originally Posted by ihavvaquestion
hey undefined...i've been thinking about this problem, and doesnt the fundamental counting principle say that we should multiply (4)(6)(8) * 6C2*4 * 4C2*6 instead of adding them on?

why do we add the last two on and not multiply if we are just counting?
We add them because the events are mutually exclusive, meaning that no two events could ever happen at the same time.

Here the events could be labeled:

A = event of choosing the bonus question such that we have done another non-bonus question in the same category.
B = event of choosing bonus question such that bonus question is MC and we did two essay questions.
C = event of choosing bonus question such that bonus question is essay and we did two MC questions.

These are three separate cases, and overall probability we are after is P(A) + P(B) + P(C).

To think of it another way. Imagine two scenarios

SCENARIO ONE

We roll a 6-sided die and flip a coin at the same time. What is the probability of rolling a 3 and getting heads?

We multiply (1/6)*(1/2) = 1/12 because the events happen at the same time.

SCENARIO TWO

We either roll a 6-sided die or we flip a coin, but we don't do both. There is a (1/2) probability of choosing to roll the die, and a (1/2) probability of choosing to flip the coin. What is the probability of rolling a 3 or getting heads?

We add (1/2)(1/6) + (1/2)(1/2) because they are separate cases.

Notice the different between and and or in red above. The multiplication rule is used for and. In more formal notation, this becomes $\displaystyle P(A \cap B)$ whereas or is $\displaystyle P(A \cup B)$.

11. that helps...thank you for clarifying