So the question is: A fair coin is tossed until a tail appears, or until it has been tossed three times. Given that the tail does not appear in the first toss, what is the conditional probability that the coin is tossed three times?
I interpreted the question as, the only way for the coin to be tossed 3 times is for it to be HHH and HHT, which would mean that the probability is 1/4.
However, is it a trick question in that, GIVEN the first toss is heads, the only probability I would need to consider is the possibility of it also being head, regardless of the first toss and the third toss, therefore, making the probability just 1/2?
Best Answer
You are failing to understand the difference between the probabilities $$P(A\ \hbox{and}\ B)\quad\hbox{and}\quad P(A\ \hbox{given}\ B)\ .$$ The probability that the first toss is heads and there are three tosses is $$P(\{HHT,HHH\})=\frac14\ ,$$ and this is what you calculated above. But the question asked is $$\eqalign{ P(\hbox{three tosses GIVEN first toss heads}) &=\frac{P(\hbox{three tosses AND first toss heads})}{P(\hbox{first toss heads})}\cr &=\frac{1/4}{1/2}\cr &=\frac12\ .\cr}$$
Comment. Conditional probabilities $P(A\ \hbox{given}\ B)$ are very important. It is definitely not a "trick question"!!!