Question:
A box contains 8 red balls, 10 white balls and 17 black balls. Two balls are drawn one by one without replacement. The probability that the first ball drawn is white and the second ball drawn is black, is
A box contains 8 red balls, 10 white balls and 17 black balls. Two balls are drawn one by one without replacement. The probability that the first ball drawn is white and the second ball drawn is black, is
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For “without replacement” problems, the total count decreases after each draw. Always update both the favourable count and the total for the second draw based on what happened in the first draw.
Updated On: Apr 25, 2026
- $\dfrac{17}{175}$
- $\dfrac{2}{35}$
- $\dfrac{5}{34}$
- $\dfrac{2}{7}$
- $\dfrac{1}{7}$
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The Correct Option is
Solution and Explanation
Step 1: Understanding the Concept:
Total balls $= 8+10+17 = 35$. Use the multiplication rule of probability for dependent events (without replacement).
Step 2: Detailed Explanation:
\[ P(\text{1st white}) = \frac{10}{35} = \frac{2}{7} \] After removing 1 white, 34 balls remain, of which 17 are black:
\[ P(\text{2nd black} \mid \text{1st white}) = \frac{17}{34} = \frac{1}{2} \] \[ P(\text{1st white and 2nd black}) = \frac{2}{7} \times \frac{1}{2} = \frac{1}{7} \]
Step 3: Final Answer:
The required probability is $\dfrac{1}{7}$.
Total balls $= 8+10+17 = 35$. Use the multiplication rule of probability for dependent events (without replacement).
Step 2: Detailed Explanation:
\[ P(\text{1st white}) = \frac{10}{35} = \frac{2}{7} \] After removing 1 white, 34 balls remain, of which 17 are black:
\[ P(\text{2nd black} \mid \text{1st white}) = \frac{17}{34} = \frac{1}{2} \] \[ P(\text{1st white and 2nd black}) = \frac{2}{7} \times \frac{1}{2} = \frac{1}{7} \]
Step 3: Final Answer:
The required probability is $\dfrac{1}{7}$.
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