Probability that a coin lands on tails an odd number of times when it is tossed $100$ times 
A coin is tossed 100 times , Find the probability that tail occurs odd number of times!

I do not know the answer, but I tried this, that there are these $4$ possible outcomes in which tossing of a coin $100$ times can unfold. 


*

*head occurs odd times

*head occurs even times

*tail occurs odd times

*tail occurs even times
Getting a head is equally likely as getting a tail, similarly for odd times and even times.
Thus, all of these events must have same the probability, i.e. $\dfrac{1}{4}$.
Is this the correct answer? Is there an alternate way of solving this problem? Lets hear it!

There are only two possible outcomes: Either both heads and tails come out an even number of times, or they both come out an odd number of times. This is so because if heads came up $x$ times and tails came up $y$ times then $x+y=100$, and the even number 100 can't be the sum of an even and an odd number.
A good way to solve this problem is to notice that if we have a sequence of 100 coin tosses in which tails came up an odd number of times, than by flipping the result of the first toss you get a sequence where tails came up an even number of times (and no matter what came up in the first toss!). Hence you have a bijection between the set of sequences where tails occurs an odd number of times, and the set of sequences where tails occurs an even number of times.
