Question

Data collected over the past few years revealed the following information: At a professional hockey teams September training camp, the probability that a player will pass all the fitness tests given he has been training all summer is 0.90. The probability that a player will not pass all the fitness tests given that he has not trained all summer is 0.45. It has been determined that four out of every five, or 80%, of players in a professional hockey teams September training camp have trained all summer. A player is randomly chosen. This player takes all the fitness tests, passing all of them. What is the probability that this player did not train all summer? In addition, write a statement interpreting the meaning of this found probability.

Answer 1

The probability, that a player has attended the training camp (Prob (A))=0. 80

The probability, that a player has not attended the training camp (Prob (Ā))= 0.20

The probability that a player will pass all the fitness tests given he has been training all summer (Prob(B/A))= 0.90

The probability that a player will not pass all the fitness tests given that he has not trained all summer (1-Prob(B/ Ā))=0.45

The probability that a player will pass all the fitness tests given that he has not trained all summer (Prob(B/Ā))= 0.55

The probability that a player selected at random clear all fitness test but did not train all summer ((Prob(Ā/B)):

                 

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There are 13% chances to select a player who will clear the fitness test given that he has not attended the hockey teams September training camp.