To solve the problem, we need to find the conditional probability that a student competed in Jive, given that they competed in Viennese Waltz.

Conditional Probability Formula:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$ Where:

• $A$ is the event that a student competed in Jive.
• $B$ is the event that a student competed in Viennese Waltz.

From the Venn diagram:

• The number of students who competed in both Jive and Viennese Waltz ($A \cap B$) is the sum of the students in the overlap of both Jive and Viennese Waltz. This includes:

  • 210 students who competed in all three dances.
  • 373 students who competed in only Viennese Waltz and Jive.

  So, $P(A \cap B) = 210 + 373 = 583$.

• The total number of students who competed in Viennese Waltz ($B$) is the sum of all students in the Viennese Waltz circle:

  • 223 (only Viennese Waltz),
  • 390 (Viennese Waltz and Bolero),
  • 373 (Viennese Waltz and Jive),
  • 210 (all three dances).

  Thus, $P(B) = 223 + 390 + 373 + 210 = 1196$.

Now, calculate the conditional probability:

$P(\text{Jive} | \text{Viennese Waltz}) = \frac{583}{1196}$

Let's compute this value.The probability that a student competed in Jive, given that they competed in Viennese Waltz, is approximately 0.487 (rounded to three decimal places).

Would you like further details or have any questions?

Here are some related questions to expand on this topic:

1. What is the probability that a student competed in all three dances?
2. What is the probability that a student competed in Bolero, given they competed in Viennese Waltz?
3. How many students competed in exactly two dances?
4. What is the probability that a randomly chosen student competed in only Jive?
5. How would the probabilities change if an additional category was added for a fourth dance?

Tip: Conditional probabilities are useful for refining predictions based on known information!