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 $80\mathrm{\%}$ chance of a high rating, a $50\mathrm{\%}$ chance if they sponsor a baseball game, and a $70\mathrm{\%}$ 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: $P(A)=0.2$, $P(B|A)=0.7$, $P(A^{\prime })=0.5$, $P(B|A^{\prime })=0.8$, $P(A^{″})=0.3$, $P(B|A^{″})=0.5$, 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: $P(B)=P(B|A)P(A)+P(B|A^{\prime })P(A^{\prime })+P(B|A^{″})P(A^{″})=0.7\ast 0.2+0.8\ast 0.5+0.5\ast 0.3=0.69$.

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

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