- #1
mynameisfunk
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Ok, This is not a homework or textbook problem; this is a question from an exam I recently took that I would like opinion on the wording of the problem.
Question: A fair coin is flipped until a head is observed and then a fair die is rolled until a 6 appears. Each flip and each roll is counted as a "trial". What is the chance that the game ends in 5 trials?
I put the culprit in bold. I interpreted the definition of a trial to be flipping a coin until you get a head and then you get to roll the die and that is the end of a trial, if you did not get a 6, then you start flipping the coin again. The reason I understood this was because of the word "and", if it had said "Each flip or each roll is counted as a trial" then I believe it would have meant the easier interpretation of just counting the ways you can flip a coin until you get a head and then you start rolling the die and never pick the coin back up.
I solved it as I interpreted it, which I may or may not be wrong, but I took the complement of the event of flipping a head and rolling a 6 which should be 1-[(1/2)(1/6)]=(11/12) then I simply said the chance of it ending exactly on the 5th trial was [(11/12)^4]*(1/12)
Sorry for the lack of LaTeX, I haven't done any in a while. Would someone please tell me their thoughts? This was a major exam and I am pretty beat up..
Question: A fair coin is flipped until a head is observed and then a fair die is rolled until a 6 appears. Each flip and each roll is counted as a "trial". What is the chance that the game ends in 5 trials?
I put the culprit in bold. I interpreted the definition of a trial to be flipping a coin until you get a head and then you get to roll the die and that is the end of a trial, if you did not get a 6, then you start flipping the coin again. The reason I understood this was because of the word "and", if it had said "Each flip or each roll is counted as a trial" then I believe it would have meant the easier interpretation of just counting the ways you can flip a coin until you get a head and then you start rolling the die and never pick the coin back up.
I solved it as I interpreted it, which I may or may not be wrong, but I took the complement of the event of flipping a head and rolling a 6 which should be 1-[(1/2)(1/6)]=(11/12) then I simply said the chance of it ending exactly on the 5th trial was [(11/12)^4]*(1/12)
Sorry for the lack of LaTeX, I haven't done any in a while. Would someone please tell me their thoughts? This was a major exam and I am pretty beat up..