Probability two balls drawn without replacement match
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A box contains $4$ red and $6$ blue balls. Two balls are drawn at random without replacement.
The probability that both are of the same colour is ________.
Show the step-by-step derivation
Step-by-step derivation
- Count the total number of unordered pairs from $10$ balls: $$\binom{10}{2} = \frac{10 \times 9}{2} = 45.$$
- Count the favourable pairs, split by colour. Both red: $$\binom{4}{2} = \frac{4 \times 3}{2} = 6.$$
- Both blue: $$\binom{6}{2} = \frac{6 \times 5}{2} = 15.$$
- The two cases are mutually exclusive, so add them: $$6 + 15 = 21\ \text{favourable pairs}.$$
- Form the probability: $$P = \frac{21}{45} = \frac{7}{15} = \mathbf{0.4667}.$$
- Cross-check by sequential probability. $$P = \underbrace{\frac{4}{10}\cdot\frac{3}{9}}_{\text{red then red}} + \underbrace{\frac{6}{10}\cdot\frac{5}{9}}_{\text{blue then blue}} = \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}.$$ ✓ The denominators drop from $10$ to $9$ precisely because the first ball is not replaced.
The idea behind this question
Drawing without replacement makes the draws dependent: once a red ball is out, the chance the next is red goes down. Multiply along the path, $P(\text{red, red}) = \frac{4}{10}\cdot\frac{3}{9}$, and add the separate cases that meet the condition. Counting combinations, $\binom{4}{2} + \binom{6}{2}$ out of $\binom{10}{2}$, gives the same answer.
Try a variation
With $3$ red and $5$ blue balls, what is the chance two drawn without replacement match?
Show the answer
Answer: $0.4643$ ($13/28$)
$\left[\binom{3}{2} + \binom{5}{2}\right] / \binom{8}{2} = 13/28$.
Other mistakes to avoid
- Adding the probabilities of the two colours when they should be multiplied along each path.
- Counting ordered pairs in the numerator and unordered pairs in the denominator.