Question:
A bag has 5 red and 7 blue balls. Two balls are drawn without replacement. What is the probability both are red?
A bag has 5 red and 7 blue balls. Two balls are drawn without replacement. What is the probability both are red?
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For probability without replacement, use sequential probabilities or combinations: $\frac{\binom{\text{favorable}}{\text{total}}}{\binom{\text{total}}{\text{chosen}}}$.
Updated On: Jul 31, 2025
- $\frac{5}{33}$
- $\frac{10}{66}$
- $\frac{5}{66}$
- $\frac{10}{33}$
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The Correct Option is A
Solution and Explanation
- Step 1: Total balls. 5 red + 7 blue = 12 balls.
- Step 2: First red ball. Probability = $\frac{5}{12}$.
- Step 3: Second red ball. 4 red, 11 total remain. Probability = $\frac{4}{11}$.
- Step 4: Joint probability. $\frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$.
- Step 5: Combinations approach. Total ways: $\binom{12}{2} = \frac{12 \cdot 11}{2} = 66$. Red ways: $\binom{5}{2} = \frac{5 \cdot 4}{2} = 10$. Probability = $\frac{10}{66} = \frac{5}{33}$.
- Step 6: Verify. Both methods give $\frac{5}{33}$, matching option (1).
- Step 7: Conclusion. Option (1) $\frac{5}{33}$ is correct.
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