Probability Without Replacement (Grades 7–8)
Sometimes one draw changes the next — like pulling marbles and not putting them back.
Understanding probability without replacement
The second probability uses the new, smaller counts. Drawing 2 red from a bag of 2 red and 3 others: P(first red) = 2/5; after taking one red, 1 red of 4 remain, so P(second red) = 1/4; both = 2/5 × 1/4 = 1/10.
Seeing it in action
Worked example
Bag: 2 red, 3 blue. Draw 2 without replacing — P(both red)?
First red: 2/5. After removing it: 1 red of 4 left → 1/4.
Multiply: 2/5 × 1/4 = 2/20 = 1/10.
The second draw uses the changed counts.
A common mistake, and the fix
Using the starting counts for the second draw.
With 2 red and 3 blue, 2/5 × 2/5 = 4/25 would be right only if the first marble went back into the bag. Without putting it back, 1 red is left among 4 marbles, so the second chance is 1/4 and both red is 2/5 × 1/4 = 1/10.
Try a few
3 red, 1 blue, draw 2 — P(both red)?
3/4 × 2/3.
4 cards, 2 aces, draw 2 — P(both aces)?
2/4 × 1/3.
5 marbles, 1 gold, draw 2 — P(gold first then any)?
1 gold of 5 on the first draw; the second draw can be anything.
Try it in Numeris.
A short lesson, then practice with feedback on every answer.
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