Not changing fractions for without replacement
"The Replacement Forget"
The Mistake in Action
A bag contains 4 red and 6 blue counters. Two counters are taken without replacement. Find the probability of getting two red.
Wrong: $P(RR) = \frac{4}{10} \times \frac{4}{10} = \frac{16}{100}$
Why It Happens
Students use the same fractions for both events, forgetting that "without replacement" means the total number changes.
The Fix
Without replacement means the first pick affects the second.
After taking one red counter:
- Red counters left: $4 - 1 = 3$
- Total counters left: $10 - 1 = 9$
$$P(RR) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$
With replacement would use $\frac{4}{10}$ twice. Without replacement — update the numbers!
Spot the Mistake
Can you identify where this student went wrong?
4 red, 6 blue. Two taken WITHOUT replacement. Find P(RR).
$P(RR) = \frac{4}{10} \times \frac{4}{10}$
Click on the line that contains the error.
Related Topics
Learn more about the underlying maths: