Step 1: Identify the three outcome categories and their probabilities: movie \(p_1 = 0.5\), play \(p_2 = 0.3\), shopping \(p_3 = 0.2\). These sum to 1, so this is a multinomial setup with \(n = 5\) trials.
Step 2: We want exactly 3 students choosing movie, 1 choosing play, and 1 choosing shopping. Use the multinomial probability formula:\[ P = \frac{n!}{n_1! \, n_2! \, n_3!} \, p_1^{n_1} p_2^{n_2} p_3^{n_3} \]with \(n_1 = 3, n_2 = 1, n_3 = 1\).
Step 3: Compute the multinomial coefficient:\[ \frac{5!}{3! \, 1! \, 1!} = \frac{120}{6} = 20 \]
Step 4: Compute the probability term:\[ p_1^3 p_2^1 p_3^1 = (0.5)^3 (0.3)(0.2) = 0.125 \times 0.06 = 0.0075 \]
Step 5: Multiply the coefficient and the probability term:\[ P = 20 \times 0.0075 = 0.15 \]So the required probability is \(\boxed{0.15}\), option (B).