Question:
A bag contains 7 green and 5 black balls. 3 balls are drawn at random one after the other. If the balls are not replaced, then the probability of all three balls being green is
A bag contains 7 green and 5 black balls. 3 balls are drawn at random one after the other. If the balls are not replaced, then the probability of all three balls being green is
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For probability without replacement, reduce both numerator and denominator for each successive draw.
Updated On: May 18, 2025
- $\dfrac{343}{1720}$
- $\dfrac{21}{36}$
- $\dfrac{12}{35}$
- $\dfrac{7}{44}$
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The Correct Option is D
Solution and Explanation
Probability that 1st ball is green = $\dfrac{7}{12}$
Probability that 2nd is green = $\dfrac{6}{11}$
Probability that 3rd is green = $\dfrac{5}{10}$
Required probability = $\dfrac{7}{12} \cdot \dfrac{6}{11} \cdot \dfrac{5}{10} = \dfrac{210}{1320} = \dfrac{7}{44}$
Probability that 2nd is green = $\dfrac{6}{11}$
Probability that 3rd is green = $\dfrac{5}{10}$
Required probability = $\dfrac{7}{12} \cdot \dfrac{6}{11} \cdot \dfrac{5}{10} = \dfrac{210}{1320} = \dfrac{7}{44}$
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