Question 600499
a) Look at this way: Let A be the event of obtaining an odd # of heads and A' be the event of obtaining an even # of heads (A' = 1 - A). Suppose you have 1 coin. Then P(A) = 1/2. If you have 2 coins, then P(A) = P(A')*(1/2) + P(A)*(1/2) = 1/2 (because we need to account for A followed by tail, or A' followed by head). 


Turns out that, by induction, P(A) = 1/2 regardless of the number of flips. Thus the probability of obtaining an odd # of heads is 1/2.


b) Similarly, we need an odd number of heads on the first 98 flips (because adding two tails won't change the parity). Hence the probability is 1/2.