The attached diagram is a map of a part of a city. Every morning, a student walks from the intersection $A$ to the intersection $B$, always only on the streets shown on the map, and always only in the north or east direction. For the sake of variety, at each intersection where a choice is possible, (independently of the other choices) with a probability of $\frac{1}{2}$, the student chooses between the two options. What is the probability that the student will pass through the intersection $C$?

![](https://cdn.mathpix.com/cropped/2024_05_02_e45df204fe230c8b1007g-1.jpg?height=927&width=895&top_left_y=360&top_left_x=607)
![](https://cdn.mathpix.com/cropped/2024_05_02_f21755f26e3475ff1a64g-1.jpg?height=794&width=1585&top_left_y=140&top_left_x=259)

Starting from $A$, we sequentially calculate the probabilities of passing through each street and intersection (see the diagram). The map is filled out by following these rules:

- The learner has a probability of 1 of touching $A$.
- If a learner passes through an intersection with probability $p$, and can proceed north and east from there, the probability of passing through the streets leading north and east from the intersection is $p / 2$.
- If a learner passes through an intersection with probability $p$, and can only proceed in one direction from there, the probability of passing through that street is also $p$.
- The probability of passing through an intersection is the sum of the probabilities of passing through the streets leading to it from the west and south.

From the diagram, it can be seen that the learner touches the $C$ intersection with a probability of $21 / 32$.

A check can be made by noting that the learner arrives at $B$ with a probability of 1, and the sum of the probabilities along the dashed lines is 1, since every path touches exactly one of the intersections lying on the same line.

Remark. It is incorrect to think that the learner's path can be described by a 7-character sequence, where three $K$ and four $E$ characters indicate whether the learner turns north or east at each intersection. There are $\binom{7}{3}=35$ such sequences. The paths that pass through $C$ are those where the three $K$ characters appear among the first six characters. There are $\binom{6}{3}=20$ such sequences, so the sought probability is 20/35. This reasoning is flawed because the sequences do not have equal probabilities. For example, the path $K K K E E E E$ has a probability of $1 / 8$, since the first three choices are made with a probability of $1 / 2$, and then the learner must always go north. In contrast, the path $E K E K E K E$ has a probability of $1 / 64$, because the first six times the learner can choose between two directions, and only on the last segment is the direction of travel certain. The formula "probability = number of favorable cases divided by the total number of cases" is only applicable when the cases are equally likely.