Question

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 37% 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

Answer 1

Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.

Conditional Probability

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

In which

• P(B|A) is the probability of event B happening, given that A happened.
• $P(A \cap B)$ is the probability of both A and B happening.
• P(A) is the probability of A happening.

In this problem:

• Event A: Fail the test.
• Event B: Unfit.

The probability of failing the test is composed by:

• 46% of 37%(are fit).
• 100% of 63%(not fit).

Hence:

$P(A) = 0.46(0.37) + 0.63 = 0.8002$

The probability of both failing the test and being unfit is:

$P(A \cap B) = 0.63$

Hence, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.63}{0.8002} = 0.7873$

0.7873 = 78.73% probability that Mona was justifiably dropped.

A similar problem is given at brainly.com/question/14398287