Im looking for an answer and I found many similar answers but the difference is that you do not know if first or second throw contains H. When I think normally if you have one H and toss then you have 1/2 probability of having T and 1/2 of another H so: HH if XH or HX (also how to simply describe such probability??). But book PROBABILITY AND RANDOM PROCESSES gives answer such as 1/3. Here is why: We have set - {HH, TH, HT, TT} Then you pull out TT cause it does not meet the condition. {HH, TH, HT} so its 1/3. But my questions are: shouldn't such set reduce to a set where order HT TH doesn't matter? -> {HH, TH} -> P(HH) = 1/2. Shouldn't we also compute it like that: 1. There is one H given. So there is 1/2 chance of it being Hx and 1/2 of xH. Also both have probability of being HH = 1/2. Hence: P(HH | Hx or xH) = 1/2 * 1/2 + 1/2 * 1/2 = 1/4+1/4 = 1/2.
Would be glad if somebody with experience checked. Thank you in advance.
