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We will calculate the total number of students and use it to find probabilities. The total number of students is the sum of all regions in the Venn diagram.

### Total number of students:
\[
378 + 359 + 318 + 214 + 371 + 374 + 312 + 203 = 2529
\]
So, the total number of students is 2529.

Now, let's compute each probability.

### a) The probability that a student competed in Merengue, given they competed in Waltz.

This is a conditional probability: \( P(\text{Merengue} \mid \text{Waltz}) \). We need to find how many students competed in both Waltz and Merengue, and then divide by the total number of students who competed in Waltz.

- Students who competed in both Waltz and Merengue: \( 214 + 371 + 374 = 959 \)
- Total students who competed in Waltz: \( 378 + 359 + 214 + 371 + 374 = 1696 \)

The conditional probability is:
\[
P(\text{Merengue} \mid \text{Waltz}) = \frac{959}{1696} \approx 0.565
\]

### b) The probability that a student competed in Paso Doble & Merengue, but not Waltz.

Here, we need to find students who competed in both Paso Doble and Merengue but not Waltz.

- This region is \( 374 \) (Paso Doble & Merengue, but not Waltz).
- Total students: 2529

The probability is:
\[
P(\text{Paso Doble and Merengue, not Waltz}) = \frac{374}{2529} \approx 0.148
\]

### c) The probability that a student competed in none of these three dances.

This is the number of students outside all three circles (region VIII), which is \( 203 \).

The probability is:
\[
P(\text{None of these dances}) = \frac{203}{2529} \approx 0.080
\]

### d) The probability that a student competed in Waltz or Merengue, but not Paso Doble.

We need to find students who competed in Waltz or Merengue but **not** in Paso Doble. This includes:
- Only Waltz: \( 378 \)
- Only Merengue: \( 312 \)
- Waltz & Merengue but not Paso Doble: \( 214 \)

So, the total number of students who competed in Waltz or Merengue but not Paso Doble is:
\[
378 + 312 + 214 = 904
\]

The probability is:
\[
P(\text{Waltz or Merengue, not Paso Doble}) = \frac{904}{2529} \approx 0.357
\]

### e) The probability that a student competed in Waltz, Paso Doble, or Merengue.

This includes all students in the three circles, so we exclude the students who competed in none of the dances (region VIII):
\[
2529 - 203 = 2326
\]

The probability is:
\[
P(\text{Waltz, Paso Doble, or Merengue}) = \frac{2326}{2529} \approx 0.920
\]

### Summary of answers:
a) \( P(\text{Merengue} \mid \text{Waltz}) = 0.565 \)  
b) \( P(\text{Paso Doble & Merengue, not Waltz}) = 0.148 \)  
c) \( P(\text{None of these dances}) = 0.080 \)  
d) \( P(\text{Waltz or Merengue, not Paso Doble}) = 0.357 \)  
e) \( P(\text{Waltz, Paso Doble, or Merengue}) = 0.920 \)

Let me know if you'd like further clarification or details! Here are some related questions:
1. What is the formula for conditional probability?
2. How do we calculate the union of two or three sets?
3. What is the principle of inclusion and exclusion in probability?
4. How can we verify that all probabilities add up to 1?
5. Can a student be in all three sets in the Venn diagram?

**Tip:** In Venn diagram problems, always label regions carefully and keep track of overlaps to avoid double-counting!

If a student is chosen at random, what is the probability that: a) The student competed in Merengue, given they competed in Waltz? b) The student competed in Paso Doble & Merengue, but not Waltz? c) The student competed in none of these three dances? d) The student competed in Waltz or Merengue, but not Paso Doble? e) The student competed in Waltz, Paso Doble, or Merengue?

Solve probability problems using a Venn diagram. Find the probability of a student competing in Waltz, Paso Doble, or Merengue. This includes conditional probability and cases with overlaps. Suitable for high school students learning basic and intermediate probability.

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