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 43% 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.)

Answer 1

Let,

F = Event that an athlete who take the test is fit enough to play soccer , P(F) = .43

Fc=Event that an athlete who take the test is not fit enough to play soccer , P(Fc) =1- .43 =.57

S= Event that an athlete pass in the fitness test.

Sc= Event that an athlete fail in the fitness test.

46% of athletes fit enough to play in the team would fail the soccer test.

P(Sc|F) = .46

So, P(S|F) = .54

No unfit athlete could possibly pass the test

P(Sc|Fc) = 1

So, P(S|Fc) = 0

Probability that Mona was justifiably dropped =P(Mona was not fit enough| Mona was dropped)

=P(Fc | Sc)= P(Sc | Fc) * P(Fc​​​​​​​) / ( P(Sc | Fc) * P(Fc​​​​​​​) + (P(Sc | F) * P (F​​​​) ) [From Bayes' theorem]

=(1*.57) / ( (1*.57)+ (.46* .43) )

=0.7423

Probability that Mona was justifiably dropped = 0.7423

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