Question

Given that we have already tossed a balanced coin ten times and obtained zero heads, what is the probability that we must toss it at least two more times to obtain the first head?

Answer 1

$Since\,\,the\,\,tosses\,\,are\,\,independent, the\,\,conditional\,\,probability\,\,equals\,\,the\,\,non-conditional\,\,probability:\\P\left( at\,\,least\,\,2 times \right) =P\left( first\,\,tail \right) =\frac{1}{2}$

The event of at least 2 tosses is that 1 toss is not enough, that means the first is tail. If the next toss is head, there are 2 tosses, otherwise more than 2 tosses.