1. **State the problem:** We want to find the probability of getting at least 4 heads when a fair coin is tossed 5 times. 2. **Formula and rules:** The number of heads in 5 tosses follows a binomial distribution with parameters $n=5$ and $p=\frac{1}{2}$ (since the coin is fair). The probability of exactly $k$ heads is given by: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ 3. **Calculate probability for at least 4 heads:** This means $P(X \geq 4) = P(X=4) + P(X=5)$. Calculate each term: - For $k=4$: $$P(X=4) = \binom{5}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{1} = 5 \times \frac{1}{16} \times \frac{1}{2} = \frac{5}{32}$$ - For $k=5$: $$P(X=5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^0 = 1 \times \frac{1}{32} \times 1 = \frac{1}{32}$$ 4. **Sum the probabilities:** $$P(X \geq 4) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16} = 0.1875$$ 5. **Round to the nearest thousandth:** $$0.1875 \approx 0.188$$ **Final answer:** The probability of getting at least 4 heads in 5 tosses of a fair coin is **0.188**.