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Statistics - Bayes theorem. Confused with the formula / how to apply it in problem solving.
Team A has played all its matches in the home away season and qualification for the finals depends on the results of three remaining matches. Teams B, C and D each have one match against three other teams and team A will make the finals if team B loses its last match and either of teams C or D also lose their last match. Sport analysis have estimated that: • Team B will lose with probability 0.8 • Team C will lose with probability 0.3 and • Team D will lose with probability 0.6 The result of any of these matches is independent of the other teams’ results and drawn matches are not possible. Based on these data, calculate the probability that team A will make the finals. Show all working.
Hope these will help: http://www.youtube.com/watch?v=XR1zovKxilw http://4.bp.blogspot.com/_wuSqJG5bIKE/TEDLT-aScwI/AAAAAAAAAjk/YIZKxuJVjxs/s1600/Bayes+Theorem.088.png
Thinking about it this is not at all a Bayes theorem problem. I think I can just solve the problem with the following: P(A)=P(B∩(C∪D)) Still, l am confused with the "and (∩)" and "or (∪)" here I'd like a confirmation please. Is ∩ means multiply the probabilities whereas ∪ means add the probabilities minus the multiplication of probabilities?
\(\cap\) (AND; probability of events intersection) is the probability that A occurs multiplied by the probability that B occurs, given that A has occurred. \[P ( A \cap B) = P(A) \times P(B \text{ } | \text{ } A)\] \(\cup\) (OR; probability of events union) is the probability that A occurs plus the probability that B occurs, minus the probability that both A and B occur. \[P ( A \cup B) = P(A) + P(B) + P ( A \cap B)\]