The problem requires calculating the probability of obtaining exactly two heads and two tails when a fair coin is tossed four times.

Step-by-Step Solution:

1. When a coin is tossed, the possible outcomes for each toss are either \(H\) (Head) or \(T\) (Tail).
2. The probability of a single outcome (head or tail) in a toss of a fair coin is \(\frac{1}{2}\).
3. Since the coin is tossed four times, the total number of possible outcomes is \((2)^4 = 16\).
4. We need to determine how many of these outcomes result in exactly two heads and two tails.
5. To find this, we use the binomial coefficient (also known as "n choose k") which determines the number of ways to choose \(k\) successes (heads) in \(n\) trials (tosses): \(\binom{n}{k}\).
6. The number of ways to get exactly two heads in four tosses is given by \(\binom{4}{2}\), which calculates as follows:

$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$

1. Therefore, there are 6 outcomes that result in exactly two heads and two tails: HH, TT, HT, HT, TH, TH.
2. Now, the probability of getting exactly two heads and two tails is the ratio of favorable outcomes to the total number of outcomes:

$\frac{6}{16} = \frac{3}{8}$

Conclusion:

Hence, the probability of obtaining exactly two heads and two tails when a fair coin is tossed four times is \(\frac{3}{8}\).

Answer Selection:

Among the given options, the correct answer is \(\frac{3}{8}\).