Answer to Question #298728 in Statistics and Probability for Travisonny

Question

One of the city's professional football teams (team 1) plays at home and another (team 2) plays away on the same night. A professional football a 0.641 probability of winning a home game and a 0.462 probability of winning an away game. Historically, when both teams play on the same night, the chance that the next morning's leading sports story will be about the team 1 is 60 percent and the chance that it will be about the team 2 game is 40 percent. Suppose that on the morning after these games the newspaper's leading sports story begins with the headline ' we win!! ' what is the probability that the story is about team 1?

Answer 1

Given

$P(A)=0.641, P(B)=0.462,$

$P(News|A)=0.6, P(News|B)=0.6$

By Bayes' Theorem

$P(A|News)=\dfrac{P(News|A)P(A)}{P(News|A)P(A)+P(News|B)P(B)}$

$=\dfrac{0.6(0.641)}{0.6(0.641)+0.4(0.462)}\approx0.675$