Immediate help needed

It is found that A and B play a game 12 times.A wins 6 times,B wins 4 times and they draw twice.A and B take part in a series of 3 games.What is the probability that they will win alternatively.

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Comments

@Kyle Finch sir ,@Brian Charlesworth sir plz help...

Tanishq Varshney - 6 years, 3 months ago

The wording of the question is a bit ambiguous. Are A and B playing just a series of 3 games, or are they playing a series of 3 matches composed of 12 games each?

If the former, then the possible sequences are ABA and BAB. Now given that A wins a game with probability 1/2, B wins a game with probability 1/3 and they draw with probability 1/6, the two sequences occur with respective probabilities (1/2)(1/3)(1/2) = 1/12 and (1/3)(1/2)(1/3) = 1/18, which add to a probability of 5/36 that they will alternate.

If they play 3 matches of 12 games each, then it gets more complicated. I would assume that winning a match entails winning 7 or more games, so there would be more sequences to analyze. I can try doing this if you think this is how the question is to be read, but at first glance I think the first interpretation is the mostly likely one.

Brian Charlesworth - 6 years, 3 months ago

Sir, why can't we consider ADB , BDA , DAB , DBA , ABD , and BAD . The winning of matches is still alternate.

Abhijeet Verma - 6 years, 2 months ago

@Abhijeet Verma – You make a good point, but I think that the default for the term "alternating" is that we are looking for sequences that only involve outright wins without any draws. Your interpretation is valid, though, so it would also need to be clarified what is meant by "alternating" ,(or "alternatively"), before this question can be resolved.

Brian Charlesworth - 6 years, 2 months ago

According to me the answer should be $11\div 36$