Why not 1/55?
P (sixth coin has 2 heads | first 5 coins get 5 heads)
=[(49/50)(48/49)(47/48)(46/47)(45/46)(1/2)^5(1/45)]/[(49/50)(48/49)(47/48)(46/47)(45/46)(1/2)^5 + 5(49/50)(48/49)(47/48)(46/47)(1/2)^4(1/46)]= 1/55
更新1:
why not ? P (sixth coin with 2 heads | 5 heads turn up in first 5 tosses) = [ P (sixth coin with 2 heads∩5 heads turn up in first 5 tosses)]/ P (5 heads turn up in first 5 tosses) = {[(49C5)/(50C5)](1/2)^5(1/45)} / {[(49C5)/(50C5)](1/2)^5 + 5[(49C4)/(50C4)](1/2)^4(1/46)} = 1/55