Maths Methods - Probability

The Roosters know that they will win 80% of their home matches and 40%
of their away matches. This season's fixture has the Roosters playing 55% of their games at home.
Given that the Roosters won their last game, what was the probability that it was played at home?

This is an application of Bayes' Theorem which deals with
conditional probabilities.
P(H)=0.55 , the probability that Roosters play at home
P(H')= the prob that Roosters do not play at home=0.45
P(W|H)=prob Roosters win given they play at home=0.80
P(W|H')=prob Roosters win given they play away=0.40

Bayes' states that P(H|W)=P(W|H)P(H)/[P(W|H)P(H)+P(W|H')P(H…
You should find that this gives 22/31 (about 71%)

probability of winning home match is 0.8
probability of playing home match is 0.55
so probability of playing home match and winning it is 0.8*0.55 = 0.44 or 44%

55% of the total matches were played at home of which 80% has the probability of winning
80%*55% = 44%