24. (2004 National College Entrance Examination Chongqing Paper) Suppose a car has to pass through 4 intersections on its way forward. The probability of the car encountering a green light (allowing passage) at each intersection is $\frac{3}{4}$, and the probability of encountering a red light (prohibiting passage) is $\frac{1}{4}$. Assume the car stops only when it encounters a red light or reaches its destination, and let $\xi$ represent the number of intersections passed when the car stops. Find: (I) the probability distribution of $\xi$ and the expectation $E \xi$;
(II) the probability that the car has passed at most 3 intersections when it stops.
24. ( I ) All possible values of $\xi$ are $0,1,2,3,4$. Let $A_{k}$ represent “the car does not stop (encounters a green light) at the $k$-th intersection”, then $P\left(A_{k}\right)=\frac{3}{4}(k=1,2,3,4)$, and $A_{1}, A_{2}, A_{3}, A_{4}$ are independent. Therefore, $P(\xi=0)=P\left(\bar{A}_{1}\right)=\frac{1}{4}, P(\xi=1)=P\left(\bar{A}_{1} \cdot A_{2}\right)=$ $\frac{3}{4} \times \frac{1}{4}=\frac{3}{16}, P(\xi=2)=P\left(A_{1} \cdot A_{2} \cdot \bar{A}_{3}\right)=\left(\frac{3}{4}\right)^{2} \cdot \frac{1}{4}=\frac{9}{64}, P(\xi=3)=P\left(A_{1} \cdot A_{2} \cdot A_{3} \cdot \overline{A_{4}}\right)=$ $\left(\frac{3}{4}\right)^{3} \cdot \frac{1}{4}=\frac{27}{256}, P(\xi=4)=P\left(A_{1} \cdot A_{2} \cdot A_{3} \cdot A_{4}\right)=\left(\frac{3}{4}\right)^{4}=\frac{81}{256}$. Thus, the distribution of $\xi$ is:
\begin{tabular}{|c|c|c|c|c|c|}
\hline$\xi$ & 0 & 1 & 2 & 3 & 4 \\
\hline$P$ & $\frac{1}{4}$ & $\frac{3}{16}$ & $\frac{9}{64}$ & $\frac{27}{256}$ & $\frac{81}{256}$ \\
\hline
\end{tabular}
$$
E \xi=0 \times \frac{1}{4}+1 \times \frac{3}{16}+2 \times \frac{9}{64}+3 \times \frac{27}{256}+4 \times \frac{81}{256}=\frac{525}{256} .
$$
(II) $P(\xi \leqslant 3)=1-P(\xi=4)=1-\frac{81}{256}=\frac{175}{256}$.