Question

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 60% chance of a high rating, a 40% chance if they sponsor a baseball game, and a 60% chance if they sponsor a professional football game. the probabilities of the company sponsoring these various games are 0.4, 0.3, and 0.3, 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)

Explanation:

Step1: Define events

Let $A$ be the event of getting a high - rating, $B_1$ be the event of sponsoring a college football game, $B_2$ be the event of sponsoring a baseball game, and $B_3$ be the event of sponsoring a professional football game. We are given $P(B_1)=0.4$, $P(B_2)=0.3$, $P(B_3)=0.3$, $P(A|B_1) = 0.6$, $P(A|B_2)=0.4$, $P(A|B_3)=0.6$.

Step2: Use the law of total probability

The law of total probability states that $P(A)=\sum_{i = 1}^{3}P(A|B_i)P(B_i)$. So $P(A)=P(A|B_1)P(B_1)+P(A|B_2)P(B_2)+P(A|B_3)P(B_3)$.
$P(A)=(0.6\times0.4)+(0.4\times0.3)+(0.6\times0.3)$
$P(A)=0.24 + 0.12+0.18$
$P(A)=0.54$.

Step3: Use Bayes' theorem

Bayes' theorem states that $P(B_3|A)=\frac{P(A|B_3)P(B_3)}{P(A)}$.
We know that $P(A|B_3) = 0.6$, $P(B_3)=0.3$ and $P(A)=0.54$.
$P(B_3|A)=\frac{0.6\times0.3}{0.54}$
$P(B_3|A)=\frac{0.18}{0.54}=\frac{1}{3}$.

Answer:

$\frac{1}{3}$