Suppose that a certain college class contains 54 students. Of these, 32 are seniors, 33 are English 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 senior and an English major?

(b) Given that the student selected is a English major, what is the probability that she is also an senior?

Let's analyze the problem step by step.

### Given Information:
- Total number of students: \( n = 54 \)
- Number of seniors: \( S = 32 \)
- Number of English majors: \( E = 33 \)
- Number of students who are neither seniors nor English majors: \( N = 7 \)

We are required to find:
(a) The probability that a student is both a senior and an English major.
(b) The probability that a student is a senior given that they are an English major.

### Step 1: Find the Number of Students who are Both Seniors and English Majors

Let \( x \) represent the number of students who are both seniors and English majors. 

We use the principle of inclusion-exclusion to find \( x \):

\[
n = S + E - x + N
\]

Substitute the given values:

\[
54 = 32 + 33 - x + 7
\]

Simplifying this equation:

\[
54 = 72 - x \quad \Rightarrow \quad x = 72 - 54 = 18
\]

So, \( x = 18 \) students are both seniors and English majors.

### (a) Probability that the Student is Both a Senior and an English Major

The probability that a randomly selected student is both a senior and an English major is given by:

\[
P(\text{Senior and English Major}) = \frac{\text{Number of students who are both seniors and English majors}}{\text{Total number of students}} = \frac{18}{54} = \frac{1}{3}
\]

### (b) Probability that the Student is a Senior Given that They are an English Major

The probability that a student is a senior given that they are an English major is a conditional probability and is given by:

\[
P(\text{Senior | English Major}) = \frac{P(\text{Senior and English Major})}{P(\text{English Major})}
\]

We already know:
- \( P(\text{Senior and English Major}) = \frac{18}{54} = \frac{1}{3} \)
- \( P(\text{English Major}) = \frac{33}{54} = \frac{11}{18} \)

Thus:

\[
P(\text{Senior | English Major}) = \frac{\frac{1}{3}}{\frac{11}{18}} = \frac{1}{3} \times \frac{18}{11} = \frac{6}{11}
\]

### Final Answers:
(a) The probability that the student is both a senior and an English major is \( \frac{1}{3} \).

(b) The probability that a student is a senior given that they are an English major is \( \frac{6}{11} \).

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Do you need more details or have any questions about this? 

Here are 8 related questions you might find interesting:

1. What is the probability that a student is a senior or an English major?
2. How many students are only seniors?
3. How many students are only English majors?
4. If two students are selected at random, what is the probability that both are English majors?
5. What is the probability that a student is neither a senior nor an English major?
6. If a student is known to be neither a senior nor an English major, what is the probability they belong to the group of 7 students?
7. How would the probabilities change if the class size increased to 60 students with the same proportions?
8. Can the concept of independent events be applied here? Why or why not?

**Tip:** In problems involving sets and probability, Venn diagrams can be a helpful tool to visualize and solve the problem.

Explore the probabilities of being both a senior and an English major in a college class of 54 students. Understand the concepts of conditional probability and set theory with detailed step-by-step solutions.

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