Independent events
P(A then B) = P(A) × P(B)
With replacement, B is independent of A: the drawn item is returned, so P(B) stays the same no matter what happened on the first draw.
Learn why each probability stays the same on every draw, then solve a tree diagram with replacement one route at a time.
With replacement means that the item you draw goes back into the collection before the next draw. The collection is fully restored, so every draw faces exactly the same probabilities.
That is why a tree diagram with replacement repeats the first set of fractions on the second branches: the draws are independent events, and no counts ever change.
P(A then B) = P(A) × P(B)
With replacement, B is independent of A: the drawn item is returned, so P(B) stays the same no matter what happened on the first draw.
Start with the full collection. Draw one branch for every possible first result, and write each probability as number of that result ÷ total items.
On every first-draw branch the drawn item goes back into the collection. The total and every category count are exactly what they were at the start.
Add the second-draw branches using the same fractions as the first draw. The branches under each parent still add up to 1.
Multiply probabilities along a complete route. When the same event can happen in several different orders, add the probabilities of all matching mutually exclusive routes.
Guided walkthrough
Use the four steps above to build the tree and solve the question one action at a time.
A bag has 3 gold and 2 silver tokens. Draw two tokens with replacement. What is the probability of one gold and one silver?
Start: 3 gold, 2 silver — 5 total.
Worked example
Problem
A bag contains 6 red marbles and 4 blue marbles. Two marbles are drawn one after another with replacement: the first marble is returned before the second draw.
Find
Find the probability of drawing at least one blue marble.
Worked example
Problem
A fair spinner has 8 equal sectors: 3 green and 5 non-green. The spinner is spun twice. Like drawing with replacement, no sector is removed, so the second spin has the same probabilities as the first.
Find
Find the probability of landing on green exactly once.
Worked example
Problem
A set of 6 letter tiles spells BANANA: 3 tiles show A and 3 show B or N. Draw two tiles one after another with replacement: the first tile goes back before the second draw.
Find
Find the probability of drawing at least one A tile.
With replacement the item goes back, so the total stays the same: the second denominator equals the first.
Because the full collection is restored, the first result cannot change the second draw. The two draws are independent events.
For one of each type, include both possible orders unless the question specifies an order, such as first then second.
For at least one target item, the complement is often simpler: find the chance of no target items, then subtract from 1.
| Topic | With replacement | Without replacement |
|---|---|---|
| Returned item | Returned to the collection | Removed from the collection |
| Second-draw denominator | Same as the first draw | One less than the first draw |
| Event relationship | Independent | Dependent |
| Calculation | Probabilities stay the same | Use probabilities conditional on the first draw |
A drawer has 5 black and 3 white pens. Take two with replacement. Find P(two black).
5/8 × 5/8 = 25/64
A card is drawn from a standard deck, returned, and drawn again. Find P(both hearts).
13/52 × 13/52 = 1/16
A jar holds 2 gold and 6 silver beads. Pick two with replacement. Find P(at least one gold).
1 − (6/8 × 6/8) = 7/16
Quick answers about independent draws and probability trees.
Divide the number of wanted objects by the total, return the drawn object, then multiply the matching path probabilities, which stay the same on every draw.
This page uses marbles, a spinner, and letter tiles to show one path, paths in either order, and a complement.
With replacement restores the collection so later events are independent. Without replacement changes the collection so later events are dependent.
The first object is returned before the second draw, so the collection is unchanged and the second set of branches repeats the first fractions.
Use the calculator to model your next probability question.