On a given weekend in the fall, a tire company can buy television and advertising time for a college football game, a baseball game, or a professional football game. If the company sponsors the college football game, there is a $80 \%$ chance of a high rating, a $50 \%$ chance if they sponsor a baseball game, and a $70 \%$ chance if they sponsor a professional football game. The probabilities of the company sponsoring these various games are $0.5,0.3$, and 0.2 , respectively. Suppose the company does get a high rating, find the probability that it sponsored a professional football game. The probability is (Type an integer or a simplified fraction.)

Step 1 ：Define the events: Let A be the event that the company sponsored a professional football game and B be the event that the company got a high rating.

Step 2 ：We are given the following probabilities: \(P(A) = 0.2\), \(P(B|A) = 0.7\), \(P(A') = 0.5\), \(P(B|A') = 0.8\), \(P(A'') = 0.3\), \(P(B|A'') = 0.5\), where A' is the event that the company sponsored a college football game and A'' is the event that the company sponsored a baseball game.

Step 3 ：Calculate the total probability of getting a high rating, \(P(B)\), using the law of total probability: \(P(B) = P(B|A)P(A) + P(B|A')P(A') + P(B|A'')P(A'') = 0.7*0.2 + 0.8*0.5 + 0.5*0.3 = 0.69\).

Step 4 ：Use Bayes' theorem to find the probability that the company sponsored a professional football game given that it got a high rating, \(P(A|B)\): \(P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{0.7*0.2}{0.69} = 0.20289855072463764\).

Step 5 ：Final Answer: The probability that the company sponsored a professional football game given that it got a high rating is approximately \(\boxed{0.203}\).