A teacher gives 6 research exercises to students, labeled a, b, c, d, e, and /. The teacher will select 3 exercises for the students to take for a grade. If a student has only studied 4 exercises, what is the probability that the student will get exactly 2 exercises right? What is the probability that the student will get at least 2 exercises right?

1. Problem Description

A teacher gives 6 research exercises to students, labeled a, b, c, d, e, and /. The teacher will select 3 exercises for the students to take for a grade. If a student has only studied 4 exercises, what is the probability that the student will get exactly 2 exercises right? What is the probability that the student will get at least 2 exercises right?

2. Solution Steps

Let N be the total number of ways to choose 3 exercises from

6. This is a combination problem, so we use the combination formula:

$N = C(n, k) = \frac{n!}{k!(n-k)!}$

where $n$ is the total number of items, and $k$ is the number of items to choose.

In our case, $n = 6$ and $k = 3$.

$N = C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

The student has studied 4 exercises and not studied 2 exercises.

a. Probability of getting exactly 2 exercises right:

To get exactly 2 exercises right, the student must choose 2 from the 4 studied and 1 from the 2 not studied.

Number of ways to choose 2 from 4 studied: $C(4, 2) = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

Number of ways to choose 1 from 2 not studied: $C(2, 1) = \frac{2!}{1!1!} = 2$

Number of ways to get exactly 2 right: $C(4, 2) \times C(2, 1) = 6 \times 2 = 12$

Probability of getting exactly 2 right: $P(\text{exactly 2}) = \frac{12}{20} = \frac{3}{5} = 0.6$

b. Probability of getting at least 2 exercises right:

"At least 2" means 2 or 3 exercises right. We already calculated the number of ways to get exactly 2 right.

Now we calculate the number of ways to get exactly 3 right.

Number of ways to choose 3 from 4 studied: $C(4, 3) = \frac{4!}{3!1!} = 4$

Number of ways to choose 0 from 2 not studied: $C(2, 0) = 1$

Number of ways to get exactly 3 right: $C(4, 3) \times C(2, 0) = 4 \times 1 = 4$

Number of ways to get at least 2 right: Number of ways to get exactly 2 right + Number of ways to get exactly 3 right = $12 + 4 = 16$

Probability of getting at least 2 right: $P(\text{at least 2}) = \frac{16}{20} = \frac{4}{5} = 0.8$

3. Final Answer

a. The probability that the student will get exactly 2 exercises right is $\frac{3}{5}$.

b. The probability that the student will get at least 2 exercises right is $\frac{4}{5}$.