24. Three people, A, B, and C, play a game: each person chooses a real number from the interval $[0,1]$, and the one whose chosen number lies between the numbers chosen by the other two wins. A randomly selects a number from the interval $[0,1]$, B randomly selects a number from the interval $\left[\frac{1}{2}, \frac{2}{3}\right]$. To maximize his winning probability, C should choose the number ( ).
(A) $\frac{1}{2}$
(B) $\frac{13}{24}$
(C) $\frac{7}{12}$
(D) $\frac{5}{8}$
(E) $\frac{2}{3}$
24. B.

Let the number chosen by C be $x$.
According to the problem, the probability that the number chosen by C is greater than the number chosen by A and less than the number chosen by B is $x\left(\frac{2}{3}-x\right)$.

Similarly, the probability that the number chosen by C is greater than the number chosen by B and less than the number chosen by A is $\left(x-\frac{1}{2}\right)(1-x)$.
Consider the function
$$
f(x)=x\left(\frac{2}{3}-x\right)+\left(x-\frac{1}{2}\right)(1-x),
$$

which is $f(x)=-2 x^{2}+\frac{13}{6} x-\frac{1}{2}$.
When $x=-\frac{\frac{13}{6}}{-2 \times 2}=\frac{13}{24}$, the probability of C winning is maximized.