Introduction
When a coin is tossed, there are two possible outcomes - heads or tails. If the coin is tossed six times, there are 2^6 = 64 possible outcomes. In this question, we are interested in finding the probability of obtaining heads and tails alternatively.

Method
To find the probability of obtaining heads and tails alternatively, we need to count the number of ways in which we can get alternate heads and tails. Let's consider the first toss. We can either get a head or a tail. Let's assume we get a head. Then, in the second toss, we must get a tail. In the third toss, we must get a head, and so on. Therefore, there are only two possible sequences that meet our requirement - HTHTHT and THTHTH.

Calculation
The probability of getting a head in the first toss is 1/2. The probability of getting a tail in the second toss is also 1/2. Therefore, the probability of getting the sequence HT is (1/2) * (1/2) = 1/4. Similarly, the probability of getting the sequence TH is also 1/4.

Since there are two possible sequences that meet our requirement, we add the probabilities to get the total probability. Therefore, the probability of obtaining heads and tails alternatively is:

P = 1/4 + 1/4 = 1/2

Conclusion
The probability of obtaining heads and tails alternatively when a coin is tossed six times is 1/2. This means that out of the 64 possible outcomes, there are 32 outcomes in which we get alternate heads and tails.