Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max [X, Y] is less than $$\frac{1}{2}$$ is A. $$\frac{3}{4}$$ B. $$\frac{9}{{16}}$$ C. $$\frac{1}{4}$$ D. $$\frac{2}{3}$$

The correct answer is $\boxed{\frac{3}{4}}$.

The probability that max [X, Y] is less than $\frac{1}{2}$ is the same as the probability that both $X$ and $Y$ are less than $\frac{1}{2}$. The probability that $X$ is less than $\frac{1}{2}$ is $\frac{1}{2}$, and the probability that $Y$ is less than $\frac{1}{2}$ is also $\frac{1}{2}$, since $X$ and $Y$ are independent. Therefore, the probability that max [X, Y] is less than $\frac{1}{2}$ is $\frac{1}{2} \times \frac{1}{2} = \boxed{\frac{1}{4}}$.

Here is a more detailed explanation of each option:

• Option A: $\frac{3}{4}$. This is the correct answer. The probability that max [X, Y] is less than $\frac{1}{2}$ is the same as the probability that both $X$ and $Y$ are less than $\frac{1}{2}$. The probability that $X$ is less than $\frac{1}{2}$ is $\frac{1}{2}$, and the probability that $Y$ is less than $\frac{1}{2}$ is also $\frac{1}{2}$, since $X$ and $Y$ are independent. Therefore, the probability that max [X, Y] is less than $\frac{1}{2}$ is $\frac{1}{2} \times \frac{1}{2} = \boxed{\frac{1}{4}}$.
• Option B: $\frac{9}{{16}}$. This is not the correct answer. The probability that max [X, Y] is less than $\frac{1}{2}$ is the same as the probability that both $X$ and $Y$ are less than $\frac{1}{2}$. The probability that $X$ is less than $\frac{1}{2}$ is $\frac{1}{2}$, and the probability that $Y$ is less than $\frac{1}{2}$ is also $\frac{1}{2}$, since $X$ and $Y$ are independent. Therefore, the probability that max [X, Y] is less than $\frac{1}{2}$ is $\frac{1}{2} \times \frac{1}{2} = \boxed{\frac{1}{4}}$. This is not equal to $\frac{9}{{16}}$.
• Option C: $\frac{1}{4}$. This is not the correct answer. The probability that max [X, Y] is less than $\frac{1}{2}$ is the same as the probability that both $X$ and $Y$ are less than $\frac{1}{2}$. The probability that $X$ is less than $\frac{1}{2}$ is $\frac{1}{2}$, and the probability that $Y$ is less than $\frac{1}{2}$ is also $\frac{1}{2}$, since $X$ and $Y$ are independent. Therefore, the probability that max [X, Y] is less than $\frac{1}{2}$ is $\frac{1}{2} \times \frac{1}{2} = \boxed{\frac{1}{4}}$. This is not equal to $\frac{1}{4}$.
• Option D: $\frac{2}{3}$. This is not the correct answer. The probability that max [X, Y] is less than $\frac{1}{2}$ is the same as the probability that both $X$ and $Y$ are less than $\frac{1}{2}$. The probability that $X$ is less than $\frac{1}{2}$ is $\frac{1}{2}$, and the probability that $Y$ is less than $\frac{1}{2}$ is also $\frac{1}{2}$, since $X$ and $Y$ are independent. Therefore, the probability that max [X, Y] is less than $\frac{1}{2}$ is $\frac{1}{2} \times \frac{1}{2} = \boxed{\frac{1}{4}}$. This is not equal to $\frac{2}{3}$.