excans:0bd71f391d

Solution: C

Let [math]T[/math] equal the number of tosses until the first head appears. [math]T[/math] has distribution [math]P(T=k) = (1/2)^k [/math] for [math]k\geq 1 [/math]. The probability that the first head appears on the fifth toss, given that it has not appeared in the first two tosses, equals

[[math]] P(T=5|T\gt2) = \frac{P(T=5)}{P(T\gt2)} = \frac{(1/2)^5}{1-0.5-0.25} = \frac{(1/2)^5}{(1/2)^2} = (1/2)^3 = 0.125. [[/math]]