Question #79228

60% of the players at a Spanish football club are local. Past experience has shown that 10% of local players have disciplinary issues and are suspended during a season, while 20% of overseas players have disciplinary issues leading to suspension. A player has just been suspended due to ill-discipline. Determine the probability that this player was from overseas

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Answer on Question #79228 – Math – Statistics and Probability

Question

60% of the players at a Spanish football club are local. Past experience has shown that 10% of local players have disciplinary issues and are suspended during a season, while 20% of overseas players have disciplinary issues leading to suspension. A player has just been suspended due to ill-discipline. Determine the probability that this player was from overseas

Solution

Let event A denote being from overseas, event B mean being suspended.

Events A and B are not independent.

Probability of being from overseas:


P(A)=10.6=0.4.P(A) = 1 - 0.6 = 0.4.


Probability of being local:


P(Aˉ)=0.6.P(\bar{A}) = 0.6.


Probability of being suspended given that the player was from overseas:


P(BA)=0.2P(B|A) = 0.2


Probability of being suspended given that the player was local:


P(BAˉ)=0.1P(B|\bar{A}) = 0.1


Probability of being from overseas and suspended:


P(AB)=P(A)P(BA)=0.40.2=0.08.P(A \cap B) = P(A) \cdot P(B|A) = 0.4 \cdot 0.2 = 0.08.


Probability of being suspended:


P(B)=P(A)P(BA)+P(Aˉ)P(BAˉ)=0.40.2+0.60.1=0.14.P(B) = P(A) \cdot P(B|A) + P(\bar{A}) P(B|\bar{A}) = 0.4 \cdot 0.2 + 0.6 \cdot 0.1 = 0.14.


Probability that the player was from overseas given the player has just been suspended


P(AB)=P(AB)P(B)=0.080.14=470.5714.P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.08}{0.14} = \frac{4}{7} \approx 0.5714.


Answer: 470.5714\frac{4}{7} \approx 0.5714.

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