1. **State the problem:** We want to find the probability of getting exactly 2 heads when a coin is tossed 4 times. 2. **Formula used:** The probability of exactly $k$ successes (heads) in $n$ independent Bernoulli trials (coin tosses) is given by the binomial probability formula: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $\binom{n}{k}$ is the binomial coefficient, $p$ is the probability of success on a single trial. 3. **Important rules:** - For a fair coin, $p = \frac{1}{2}$. - The binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ counts the number of ways to choose $k$ successes out of $n$ trials. 4. **Apply values:** Here, $n=4$, $k=2$, and $p=\frac{1}{2}$. Calculate the binomial coefficient: $$\binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$$ 5. **Calculate probability:** $$P(X=2) = 6 \times \left(\frac{1}{2}\right)^2 \times \left(1 - \frac{1}{2}\right)^{4-2} = 6 \times \frac{1}{4} \times \frac{1}{4}$$ 6. **Simplify:** $$6 \times \frac{1}{4} \times \frac{1}{4} = 6 \times \frac{1}{16} = \frac{6}{16}$$ 7. **Reduce fraction:** $$\frac{6}{16} = \frac{\cancel{2} \times 3}{\cancel{2} \times 8} = \frac{3}{8}$$ **Final answer:** The probability of getting exactly 2 heads in 4 tosses is $\boxed{\frac{3}{8}}$.