Problem 20. Five bricks are placed on the ground as shown. Peter can only remove a brick if there are no bricks on top of it. He selects one of the available bricks at random and removes it until all the bricks are removed. What is the probability that the brick numbered 4 is the third brick to be removed? So, in order to be able to remove the brick numbered 4, we have to first remove bricks 1 and 2. And so if brick 4 is the third brick to be removed, that means that bricks 1 and 2 are the first and second bricks to be removed in some order. It doesn't matter which. So, in our first move, the only possibilities are to either remove bricks 1 or 2 because everything else has something on top of it. And either of those gets us towards the goal of brick 4 being the third brick to remove. So it doesn't matter what we do in the first move. Now, in our second move, we have two choices. Assume first that we remove brick 1 in the first move. Then we'll either be able to remove brick 2 or we'll be able to remove brick 3, which will now be uncovered now that brick 1 is gone. But bricks 4 and 5 are not removable because they're still covered by brick 2. If instead we'd removed brick 2 in the first round, the situation would be similar. We'd be able to remove either brick 1 or brick 5, but not bricks 3 and 4. If we remove either brick 3 or brick 5, then we won't be able to remove brick 4 in the third move because there will still be one of the two bricks on the top row that is still there. So we have to remove the brick on the top row, which is one out of the two possibilities, no matter what we've removed in the first move. So there's a probability half that we choose the right brick to remove. Now, given that we've removed the correct two bricks in the first two moves, we now have just the bottom row. And we can remove any of those three bricks. And there's a 1 third probability that we will pick brick 4 as the brick to remove. So in total, there's a probability of 1 that we're on track after the first move. Then given that there's a half probability that we pick the right thing in the second move, and then a 1 third probability that we pick the right thing in the third move. So a total probability that we remove brick 4 as the third brick is 1 sixth. And so the answer is D.

probability

brick removal

stack of bricks

combinatorics

probability calculation