$\frac{3}{8}$
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:
- When a coin is tossed, the possible outcomes for each toss are either \(H\) (Head) or \(T\) (Tail).
- The probability of a single outcome (head or tail) in a toss of a fair coin is \(\frac{1}{2}\).
- Since the coin is tossed four times, the total number of possible outcomes is \((2)^4 = 16\).
- We need to determine how many of these outcomes result in exactly two heads and two tails.
- 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): \(\binom{n}{k}\).
- The number of ways to get exactly two heads in four tosses is given by \(\binom{4}{2}\), which calculates as follows:
$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$
- Therefore, there are 6 outcomes that result in exactly two heads and two tails: HH, TT, HT, HT, TH, TH.
- 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 \(\frac{3}{8}\).
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