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Probability with and without replacement

20 practice questions 2 video lessons Theory + worked examples

Probability With and Without Replacement

Common Core Geometry • Standard S-CP.8 • Probability & Statistics

Probability With and Without Replacement is a topic in Probability & Statistics in the Common Core State Standards. It is aligned to Standard S-CP.8, which requires students to apply the general multiplication rule and compute probabilities with and without replacement.

With replacement the pool is unchanged and draws are independent; without replacement the counts drop, making draws dependent.

Common Core Geometry › Probability & Statistics › Probability With and Without Replacement  —  Standard S-CP.8

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Practice questions

Every question with a fully worked solution.

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  • 2 Examples of Probability With & Without Replacement Watch
  • Introduction to Probability With and Without Replacement Watch
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Theory

When drawing several items, replacement decides whether the pool changes:

  • With replacement: the item is returned, so probabilities stay the same and the draws are independent.
  • Without replacement: the item is kept out, so the total (and that count) drop by one, and the draws are dependent.
Multiply along the sequence of draws, updating the probabilities after each draw if there is no replacement.
With or without replacement Without replacement, the second draw has one fewer item, so its probabilities change. 3/5 2/5 Red Blue 1st draw (3 red, 2 blue) without replacement: 2nd draw uses 4 left
Without replacement, the second draw uses one fewer item.
Replacement Replacement Replacement with replacement: probabilities stay the same without replacement: total (and count) drop by 1 multiply along the branch of draws
With vs without replacement.

Multiply the (updated) probabilities:

\[P(\text{both})=P(\text{first})\times P(\text{second}\mid\text{first})\]
the probability of both draws is the first times the conditional probability of the second
Without replacement, the second factor is conditional — the denominator is one smaller.

How to handle replacement

  1. Decide whether items are replaced.
  2. First draw: use the original counts.
  3. Second draw: keep counts (with) or reduce them (without).
  4. Multiply the probabilities along the branch.
Example 1 — Without replacement
A bag has \(3\) red and \(2\) blue. Draw two without replacement. Find \(P(\text{both red})\).
Solution

After a red is drawn, \(2\) red of \(4\) remain.

\(P(RR)\)\(=\)\(\dfrac35\cdot\dfrac24\)
\(=\)\(\dfrac{6}{20}=\dfrac{3}{10}\)
the probability is three tenths
Example 2 — With replacement
Same bag, but replace the first draw. Find \(P(\text{both red})\).
Solution

With replacement the counts reset, so the draws are independent.

\(P(RR)\)\(=\)\(\dfrac35\cdot\dfrac35\)
\(=\)\(\dfrac{9}{25}\)
the probability is nine twenty-fifths
Example 3 — Different colors
From the same bag, draw two without replacement. Find \(P(\text{red then blue})\).
Solution

Red first, then blue from the \(4\) remaining.

\(P(RB)\)\(=\)\(\dfrac35\cdot\dfrac24\)
\(=\)\(\dfrac{6}{20}=\dfrac{3}{10}\)
the probability is three tenths
Example 4 — Why it changes
Why does the second probability change without replacement?
Solution

Because an item is removed, so both the favorable count and the total decrease — the draws are dependent.

removing an item changes the counts, making the draws dependent

Common pitfalls

Without replacement, reduce the total for the next draw.
With replacement the draws are independent; without, they are dependent.
Update the favorable count too, not just the total, when drawing the same color again.

Frequently asked questions

What does “with replacement” mean?

The drawn item is returned, so the probabilities stay the same and draws are independent.

What does “without replacement” mean?

The item is kept out, so the total drops by one and the draws are dependent.

How do you compute probability over several draws?

Multiply the probability of each draw, updating counts after each if there is no replacement.

Why are draws without replacement dependent?

Because removing an item changes the counts, so the earlier draw affects the later probability.