Probability: Edge Cases and Subtle Traps (5)

Question

A bag contains $5$ red balls and $3$ blue balls. Two balls are drawn one after the other without replacement. Find the probability that both balls are red.

The trap here: students confuse “without replacement” with “with replacement” and use the wrong probability for the second draw. Let’s do this carefully.

Solution — Step by Step

Why This Works

The key idea: in conditional probability, the sample space changes after the first event. When you remove a red ball, the remaining bag is different, so the probability of drawing red again is calculated from the new distribution.

Multiplication rule: $P(A \cap B) = P(A) P(B|A)$. For independent events, $P(B|A) = P(B)$ and the rule simplifies. Here, the events are dependent — drawing without replacement creates dependence.

Alternative Method

Counting directly. Number of ways to pick $2$ red from $5$: $\binom{5}{2} = 10$. Number of ways to pick any $2$ from $8$: $\binom{8}{2} = 28$.

$P = \frac{10}{28} = \frac{5}{14} \checkmark$

(Note: combinations work because order doesn’t matter for “both red.” If we’d computed using permutations, $P = (5 \times 4)/(8 \times 7) = 20/56 = 5/14$ — same.)

Common Mistake

The classic trap: students write $P(\text{both red}) = (5/8)^2 = 25/64$, treating the two draws as independent. That’s only correct with replacement. Without replacement, the second draw’s denominator drops to $7$ and the numerator to $4$ (one red is gone).

A subtler trap: in conditional probability problems like “given the first is red, find the probability the second is red,” students sometimes forget to update both numerator AND denominator. Both must reflect the new sample.