Probability

Tree diagrams: with replacement vs without replacement

By Mr Gan · O-Level E Maths · Updated August 2026 · 7 min read

Building on the sample space and complement rule, a probability tree diagram shows every possible outcome of two or more stages as branches: multiply the probabilities along a branch to get that outcome's chance, and add the results of branches that match the question. The key distinction is whether the draw is with replacement (the object goes back in, so the denominators on the second draw stay the same) or without replacement (the object stays out, so the denominators on the second draw shrink by one, and the numerators may shrink too).

Why the second branch trips students up

The tree diagram itself is not hard to draw. Where marks disappear is the second set of branches. Students copy the first-draw probabilities straight across to the second draw without checking whether the item was replaced. In a "with replacement" question this is correct, because the total number of items is the same for both draws. In a "without replacement" question it is wrong, because one item has already left the bag, box, or drawer, and both the total and possibly the count of the colour just drawn have changed.

Get into the habit of reading the question twice: once for what is being counted, and once for the single word "replacement" or "without replacement". That word decides whether your second set of branches is identical to the first or different.

The method

Step 1

Draw branches for each stage, label each with its probability

Step 2

Branches from the same point must add to 1

Step 3

Multiply along a path for that path's probability

Step 4

Add every path that matches the outcome asked for

Watch the whole method in about a minute.

Worked example 1: with replacement

A bag contains 5 red balls and 3 blue balls. A ball is drawn, its colour noted, and it is put back in the bag before a second ball is drawn. Find P(both balls are the same colour).

Solution

There are 8 balls in total. Because the first ball is replaced, the second draw has the same 8 balls: P(red) = 5/8 and P(blue) = 3/8 on both draws.

"Same colour" means either both red or both blue. Multiply along each of these two paths:
P(RR) = 5/8 × 5/8 = 25/64
P(BB) = 3/8 × 3/8 = 9/64

Add the two paths, since either one satisfies "same colour":
P(both same colour) = 25/64 + 9/64 = 34/64 = 17/32

Quick check: the four outcomes RR, RB, BR, BB must have probabilities that add to 1. Here 25/64 + 15/64 + 15/64 + 9/64 = 64/64 = 1, so the working is consistent.

Worked example 2: without replacement

The same bag has 5 red balls and 3 blue balls. This time a ball is drawn and not replaced before the second ball is drawn. Find P(exactly one red ball).

Solution

First draw, 8 balls: P(red) = 5/8, P(blue) = 3/8.

Second draw only has 7 balls left, and the count of each colour depends on what the first draw was. If the first ball was red: 4 red and 3 blue remain, so P(blue | red first) = 3/7. If the first ball was blue: 5 red and 2 blue remain, so P(red | blue first) = 5/7.

"Exactly one red" is satisfied by two paths: red-then-blue, or blue-then-red. Multiply along each:
P(R then B) = 5/8 × 3/7 = 15/56
P(B then R) = 3/8 × 5/7 = 15/56

Add the two paths:
P(exactly one red) = 15/56 + 15/56 = 30/56 = 15/28

The step students get wrong

Writing 3/8 again for the second-draw blue probability, instead of 3/7. Without replacement, the denominator drops by 1 on every later draw (one fewer ball in the bag), and if the colour drawn first matches the colour you are now finding, its numerator drops by 1 too. Always ask: how many balls are left, and how many of the colour I want are left, given what happened on the branch above.

Frequently asked questions

How is a tree diagram different from a Venn diagram for probability?

A tree diagram is best when events happen in stages, one after another, such as two draws from a bag. A Venn diagram is best when you are comparing overlap between two or more categories that happen at the same time. See Venn diagrams and set notation for the overlap case.

Do I always multiply along the branches and add across paths?

Yes. Multiplying along a path gives the probability that a specific sequence of outcomes happens (this uses the same idea as independent and conditional events). Adding across paths combines separate sequences that all count as the outcome the question is asking for, because the paths cannot happen at the same time.

How do I know whether to treat two draws as independent?

With replacement, the two draws are independent: the second draw's probabilities do not depend on the first. Without replacement, they are not independent, and the second draw's probabilities depend on what was drawn first. See mutually exclusive vs independent events for the full distinction.

Do I need to draw the full tree, or can I just calculate?

For O-Level E-Maths, drawing the tree with labelled branches is usually expected working, even if the final calculation is short. It also makes it far easier to spot which paths you need to add, and to check your branch probabilities sum to 1 at each stage.

— Mr Gan Math Tuition

Mixing up with and without replacement?

Mr. Gan works with students until tree diagrams stop being a guessing game and become a checklist you can run under exam pressure.

Chat with Mr. Gan