A manager decides to distribute Rs. 20000 between two employees X and Y. He knows X deserves more than Y, but does not know how much more. So, he decides to arbitrarily break Rs. 20000 into two parts and gives X the bigger part. Then, the chance that X gets twice as much as Y or more is
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- $\dfrac{2}{5}$
- $\dfrac{1}{2}$
- $\dfrac{1}{3}$
- $\dfrac{2}{3}$
The Correct Option is D
Solution and Explanation
We are to find the probability that $20000 - y \ge 2y \Rightarrow 20000 \ge 3y \Rightarrow y \le \dfrac{20000}{3} \approx 6666.67$
Since X must get more, $y$ ranges from $0$ to $10000$
Favorable range for $y$: $0 \le y \le \dfrac{20000}{3}$
Total range: $0 \le y<10000$
Required probability = $\dfrac{20000/3}{10000} = \dfrac{2}{3}$
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