Suppose that a certain college class contains 63 students. Of these, 33 are juniors, 37 are chemistry majors, and 7 are neither. A student is selected at random from the class.

(a) What is the probability that the student is both a junior and a chemistry major?

(b) Given that the student selected is a chemistry major, whet is the probability that she is also a junior?

Let's solve this step by step.

### Given:
- Total number of students in the class, \( n = 63 \).
- Number of juniors, \( |J| = 33 \).
- Number of chemistry majors, \( |C| = 37 \).
- Number of students who are neither juniors nor chemistry majors, \( |N| = 7 \).

### Step 1: Determine the number of students who are either juniors or chemistry majors.
Since 7 students are neither juniors nor chemistry majors, the number of students who are either juniors or chemistry majors is:

\[
|J \cup C| = n - |N| = 63 - 7 = 56
\]

### Step 2: Use the principle of inclusion-exclusion to find the number of students who are both juniors and chemistry majors.
The formula for the union of two sets is:

\[
|J \cup C| = |J| + |C| - |J \cap C|
\]

Where \( |J \cap C| \) is the number of students who are both juniors and chemistry majors. Rearranging the equation to solve for \( |J \cap C| \):

\[
|J \cap C| = |J| + |C| - |J \cup C| = 33 + 37 - 56 = 14
\]

So, there are 14 students who are both juniors and chemistry majors.

### Part (a): Probability that the student is both a junior and a chemistry major.

The probability is the ratio of students who are both juniors and chemistry majors to the total number of students:

\[
P(\text{Junior and Chemistry Major}) = \frac{|J \cap C|}{n} = \frac{14}{63} = \frac{2}{9}
\]

### Part (b): Given that the student selected is a chemistry major, what is the probability that she is also a junior?

This is a conditional probability, calculated as:

\[
P(\text{Junior} \mid \text{Chemistry Major}) = \frac{P(\text{Junior and Chemistry Major})}{P(\text{Chemistry Major})} = \frac{|J \cap C|}{|C|} = \frac{14}{37}
\]

### Final Answers:
(a) The probability that the student is both a junior and a chemistry major is \(\frac{2}{9}\).

(b) Given that the student selected is a chemistry major, the probability that she is also a junior is \(\frac{14}{37}\).

Would you like to see more details or have any questions?

Here are some related questions you might be interested in:
1. How would the probability change if there were 10 students who were neither juniors nor chemistry majors?
2. What is the probability that a randomly selected student is a junior or a chemistry major?
3. If the class size were increased to 100 students, how might the probabilities be affected?
4. Can we calculate the probability that a student is a junior given that they are not a chemistry major?
5. How would these probabilities be represented using Venn diagrams?
6. What if the number of chemistry majors is greater than the number of juniors? How does that affect the calculation?
7. How can we calculate the probability if the conditions are switched (given that the student is a junior, what is the probability that they are a chemistry major)?
8. What assumptions are we making in these probability calculations?

**Tip:** When dealing with probability problems, clearly identify and differentiate between "and," "or," and "given that" (conditional) probabilities to avoid confusion.

Explore the probability of a student being both a junior and a chemistry major in a college class of 63 students. Understand the application of conditional probability and the inclusion-exclusion principle. Solutions and step-by-step calculations provided.

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