Any Athlete Who Fails State University's Women's Team

Any Athlete Who Fails The Enormous State Universitys Womens Soccer F

Any athlete who fails the Enormous State University's women's soccer fitness test is automatically dropped from the team. Last year, Mona Header failed the test, but claimed that this was due to the early hour. (The fitness test is traditionally given at 5 AM on a Sunday morning.) In fact, a study by the ESU Physical Education Department suggested that 46% of athletes fit enough to play on the team would fail the soccer test, although no unfit athlete could possibly pass the test. It also estimated that 39% of the athletes who take the test are fit enough to play soccer. Assuming these estimates are correct, what is the probability that Mona was justifiably dropped? (Round your answer to four decimal places.)

Paper For Above instruction

The question revolves around assessing the probability that Mona Header was justifiably dropped from the women's soccer team at Enormous State University (ESU) due to failing the fitness test. To analyze this, we employ conditional probability principles, specifically Bayes' theorem, to determine the probability that Mona was unfit (and thus justifiably dropped) given her failure of the test.

Understanding the Data and Setting Up the Problem

The information provided includes:

1. No unfit athlete can pass the test.

2. 46% of athletes who are fit enough to play on the team would fail the test.

3. 39% of athletes taking the test are fit enough to play.

4. Mona failed the test, and she claims it was due to the early hour.

Our aim is to find the probability that Mona was unfit given she failed the test, denoted as \( P(\text{Unfit} | \text{Failed}) \).

Defining the Events:

- \( F \): Athlete is fit enough to play soccer.

- \( U \): Athlete is unfit.

- \( T \): Athlete fails the test.

From the problem, we know:

- \( P(F) = 0.39 \) (probability an athlete is fit)

- \( P(U) = 1 - P(F) = 0.61 \)

Since no unfit athlete can pass the test:

- \( P(\text{Pass} | U) = 0 \)

- Therefore, \( P(T | U) = 1 \) (unfit athletes always fail).

Likewise, among fit athletes:

- \( P(T | F) = 0.46 \) (46% of fit athletes fail)

- \( P(\text{Pass} | F) = 1 - 0.46 = 0.54 \) (54% of fit athletes pass).

Applying Bayes' Theorem

The probability that Mona was unfit given she failed the test:

\[

P(U | T) = \frac{P(T | U) \times P(U)}{P(T)}

\]

First, we need \( P(T) \), the total probability that an athlete fails the test. Using the Law of Total Probability:

\[

P(T) = P(T | F) \times P(F) + P(T | U) \times P(U) = (0.46)(0.39) + (1)(0.61)

\]

Calculating:

\[

P(T) = 0.1794 + 0.61 = 0.7894

\]

Thus,

\[

P(U | T) = \frac{(1)(0.61)}{0.7894} \approx \frac{0.61}{0.7894} \approx 0.7722

\]

Interpretation

The probability that Mona was unfit (and thus justifiably dropped) given she failed the test is approximately 0.7722, or 77.22%. This high probability indicates that Mona's failure was likely due to her unfitness rather than external factors like the early hour.

Conclusion

Based on the data and the assumptions, there is approximately a 77.22% chance that Mona was justifiably dropped from the team because she was unfit to play, as indicated by her failing the fitness test.

References

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