I’m trying to wrap my head around tree diagrams for probability, and I keep tripping over what changes on the second level of branches. Say I’m picking two marbles from a bag without replacement and I need the probability that the first is red given that the second is blue. I know I’m supposed to draw a tree, but I get stuck on two things:
– After the first branch (e.g., first pick is red or blue), do the second-branch probabilities have to be different on each side, or can they be the same? I feel like they should change because the bag has changed, but then I second-guess myself and wonder if I’m double-counting something.
– When the question says “given that the second pick is blue,” do I cross out all branches where the second pick isn’t blue and then renormalize the remaining branches, or do I need to redraw a new tree from scratch that starts with the second pick being blue?
Also, if the problem were with replacement instead, would the second-level probabilities just be identical to the first level on both sides of the tree, or is there still some subtlety I’m missing? I might be overthinking this, but I keep getting tangled. Any tips on the right way to set up the tree for these cases?
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3 Responses
You’re thinking about it exactly the right way: without replacement, the second-level probabilities do change because the bag has changed. If you start with r red and b blue, then from the “first is red” branch the chance the second is blue is b/(r+b−1), while from the “first is blue” branch it’s (b−1)/(r+b−1)-different because you removed different colors. For “given that the second pick is blue,” you don’t need a new tree; keep your original tree, cross out the branches where the second isn’t blue, and renormalize what’s left. Algebraically, P(first red | second blue) = P(R1 ∩ B2) / P(B2) = [r/(r+b) · b/(r+b−1)] / {[r/(r+b) · b/(r+b−1)] + [b/(r+b) · (b−1)/(r+b−1)]} = r/(r+b−1). Intuition: once you know the second is blue, the first is just a random pick from the remaining r+b−1 marbles, r of which are red. With replacement, the second-level probabilities are identical on both sides (the bag “resets”), so the draws are independent and P(first red | second blue) = P(first red) = r/(r+b). I used to redraw trees from scratch and confuse myself; a tutor showed me the “slice and renormalize” trick, and it finally clicked-I even started color-coding the “given” branches. A nice walkthrough is here: https://www.khanacademy.org/math/statistics-probability/probability-library/conditional-probability-independence/v/conditional-probability-tree-diagram-example.
– Without replacement, the second-branch probabilities change on each side (the counts changed), while with replacement they’re identical to the first level on both sides because the draws are independent.
– For “given the second is blue,” keep only the branches where the second is blue and renormalize-i.e., P(first red | second blue) = P(first red ∩ second blue)/P(second blue); no need to redraw the tree (see https://www.khanacademy.org/math/statistics-probability/probability-library/tree-diagrams/v/tree-diagrams).
Think of the bag as starting with R red and B blue marbles (N = R + B). Without replacement, the second-level probabilities depend on what happened first, because the counts change. On the branch “first red,” the remaining bag has R−1 red and B blue, so P(second blue) = B/(N−1); and (by symmetry) many people write P(second red) = R/(N−1) on that branch too, even though the bag has one fewer red than it started with. On the branch “first blue,” the remaining bag has R red and B−1 blue, so P(second blue) = (B−1)/(N−1) and P(second red) = R/(N−1). With replacement, the counts reset each time, so the second-level probabilities are the same as the first level on both sides of the tree.
For “given that the second pick is blue,” you don’t need a new tree. Take the full two-level tree, cross out every path whose second pick isn’t blue, and then renormalize what’s left so those remaining probabilities sum to 1. The conditional probability P(first red | second blue) is the weight on the path “first red then blue” divided by the total weight of all “second blue” paths. A neat shortcut here: conditioning on “second is blue” is like reserving one blue for the second draw; the first draw then comes from N−1 marbles with R red and B−1 blue, so P(first red | second blue) = R/(N−1).
With replacement, the draws are independent, so conditioning on the second being blue doesn’t change the first: P(first red | second blue) = P(first red) = R/N. You can still do the cross-out-and-renormalize step if you like, but because the branch probabilities don’t change from level to level, the numbers line up the same way. Hope this helps!