What we see: Problem Solving, probability. Jane walks 3 blocks south (S) and 3 blocks east (E) in some order, every order equally likely, and we want the chance that her first 2 blocks are both south.
Step 1. Count all the paths.
What we see: every path is 6 blocks long, with 3 S and 3 E.
What it means: a path is set the moment we choose which 3 of the 6 spots are S. The other 3 spots are E.
What we do and why: we count those choices, because each one is a different path and each is equally likely. \( \binom{6}{3} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20 \) paths.
Step 2. Count the paths that start S, S.
What we see: if the first 2 blocks are S, there are 4 blocks left: 1 S and 3 E.
What it means: the only choice left is where that last S goes among the 4 spots.
What we do and why: we count those spots. \( \binom{4}{1} = 4 \) paths.
Step 3. Divide.
What we see: 4 good paths out of 20.
What it means: probability is good paths divided by all paths.
What we do and why: \( \frac{4}{20} = \frac{1}{5} \).
The trap: C, \( \frac{2}{5} \). That is the chance the second block is south once the first block already was south. It forgets to also multiply by the chance that the first block is south (\( \frac{1}{2} \)). The other trap is treating each block like a coin flip, \( \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \). That is wrong because Jane has only 3 south blocks: once she uses one, fewer are left.
Faster way: The first block is south with chance \( \frac{3}{6} \). Then 2 of the 5 blocks left are south, so the second is south with chance \( \frac{2}{5} \). Multiply: \( \frac{3}{6} \cdot \frac{2}{5} = \frac{6}{30} = \frac{1}{5} \). Under 30 seconds.
Takeaway: When you pick from a fixed set without putting anything back, the chances change after each pick. Multiply the chance of each step, using one fewer item each time.
Answer: B
