Problem Solving

In a high school debating team consisting of 2 freshmen, 2 sophomores, 2 juniors, and 2 seniors, two students are selected to represent the school at the state debating championship. The rules stipulate that the representatives must be from different grades, but otherwise the 2 representatives are to be chosen by lottery. What is the probability that the students selected will consist one freshman and one sophomore?

A. 1/16
B. 1/8
C. 1/7
D. 1/6
E. 1/4

OA: D

Resurgent wrote:In a high school debating team consisting of 2 freshmen, 2 sophomores, 2 juniors, and 2 seniors, two students are selected to represent the school at the state debating championship. The rules stipulate that the representatives must be from different grades, but otherwise the 2 representatives are to be chosen by lottery. What is the probability that the students selected will consist one freshman and one sophomore?

A. 1/16
B. 1/8
C. 1/7
D. 1/6
E. 1/4

OA: D

Let F be the event that the first student selected is Freshman. And S be the event the 2nd student is Sophomore. These two are equally likely so we will multiply our result by 2. We know that

P(S|F) = P( S and F)/P(F). We are conditioning on the fact that a freshman was chosen first. Probability of choosing a Freshman or a sophomore first is 2/8 =1/4
P( S and F)= P(F) x P(S|F)
= 1/4 x P(S|F)
This is the tricky part. If a Freshman has already been taken then we are left with 7 students over all. But the restriction says No other freshman can be chosen so the remaining freshman cannot be included in the sample space in the calculating P(S|F). So the prob that the second student is a Sophomore is 2/6 =1/3
P( S and F)= 1/4 x 1/3 =1/12

If we reverse roles we get the same value 1/12
1/12 + 1/12 = 1/6

This is an intriguing and interesting question.

(2*2)/(7*8/2) = 1/7

F F So So J J Se Se

Choosing either Freshmen first = 2*6 (two freshmen to choose from and six remaining possibilities)

Repeat for each class gives us 48. However, need to divide by 2 so that we dont double count. (picking freshmen first and then sophmore is the same as the reverse).

So a total of 24 possibilities.

Next step is to find F and So pairings, so distinguish b/t the 4.
F1 & So1
F1 & So2
F2 & So1
F2 & So2

4/24 = 1/6

are you sure the answer is not (1/7)?

Consider them 4 team with 2 people in each of them.
Now, we need to select 2 people, each from one of these 4 teams so,
4C2 = 6.

So, 6 combination exists with 2 people from different teams and only 1 combination of Fresher and Sophomore is required.

So, effectively, 1/6

The way I solved it was:

There's a total of 8 students to choose from. First lets choose one freshman or sophomore. There's a total of 4 freshmen and sophomores. So the probability that you can choose either one freshman or sophomore out of a total of 8 students is 4/8. Lets say we effectively get a freshman. That leaves us with 6 students left that we can choose from because we got our freshman. Now the probability that we'll choose 1 sophomore out of 6 students is 2/6. Since we're looking for one freshman AND one sophomore, we have to multiply the two probability leaving us with:

(4/8)*(2/6)=8/48 which equals 1/6.

Hope that makes sense