A manager decides to distribute Rs. 20000 bet | vedclass

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(D) Let the total amount be $S = 20000$. Let $X$ be the amount given to employee $X$ and $Y$ be the amount given to employee $Y$. Given $X + Y = 20000

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$2/3$

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(D) Let the total amount be $S = 20000$. Let $X$ be the amount given to employee $X$ and $Y$ be the amount given to employee $Y$. Given $X + Y = 20000$ and $X > Y$. Since $X$ is the bigger part,$X$ must be in the range $(10000, 20000]$. The total length of the sample space for $X$ is $20000 - 10000 = 10000$. We want the probability that $X \geq 2Y$. Substituting $Y = 20000 - X$,we get $X \geq 2(20000 - X)$. $X \geq 40000 - 2X \Rightarrow 3X \geq 40000 \Rightarrow X \geq 40000/3$. Since $X$ must be at most $20000$,the favorable range for $X$ is $[40000/3, 20000]$. The length of this interval is $20000 - 40000/3 = 20000/3$. The probability is $\frac{20000/3}{10000} = \frac{2}{3}$.

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