A fair coin is flipped four times. Use a tree diagram to find the probability of observing exactly two heads. Enter your answer as a percentage, to the nearest percent.

To find the probability of getting exactly two heads when flipping a fair coin four times, we first consider the sample space. A tree diagram would show all possible outcomes (16 in total). However, we can simplify our calculations with a combination approach. The number of ways to choose 2 heads from 4 flips can be calculated using the binomial coefficient: \[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] Each sequence of flips (like HHTT, HTHT, etc.) has a probability of \( (0.5)^2 \times (0.5)^2 = 0.5^4 = \frac{1}{16} \). Now, multiply the number of favorable outcomes by the probability of each: \[ P(\text{exactly 2 heads}) = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8} \] To express this as a percentage: \[ \frac{3}{8} = 0.375 \times 100\% = 37.5\% \] Therefore, to the nearest percent, the probability of observing exactly two heads is **38%**.