The correct answer is $\boxed{\frac{1}{2}}$.
A fair coin has a probability of $\frac{1}{2}$ of landing on heads and a probability of $\frac{1}{2}$ of landing on tails. The probability of getting four heads in a row is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. However, this does not mean that the probability of getting a tail on the next toss is $\frac{15}{16}$. The probability of getting a tail on the next toss is still $\frac{1}{2}$, regardless of the outcome of the previous tosses.
To understand this, it is helpful to think about the coin toss as a random event. Each time the coin is tossed, the outcome is independent of the previous outcomes. This means that the probability of getting a tail on the next toss is not affected by the fact that the previous four tosses were heads.
Another way to think about this is to consider the following analogy. Suppose you have a bag of marbles that contains 10 red marbles and 10 blue marbles. You reach into the bag and pull out a red marble. What is the probability that the next marble you pull out will be red? The answer is still $\frac{1}{2}$, even though the previous marble was red. This is because the probability of getting a red marble is not affected by the color of the previous marble.
In conclusion, the probability of obtaining a ‘Tail’ when the coin is tossed again is $\frac{1}{2}$.