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
Answer
0.4667 i.e. 7/15 (accept 0.46 to 0.47) i.e. 7/15 (accept 0.46 to 0.47)
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 trap this question is built around
Using $ (4/10)^2 + (6/10)^2 = 0.52 $, which silently assumes replacement. Without replacement the second draw is conditioned on the first, and the answer is lower - drawing one ball makes its colour scarcer.