Question:
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
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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When a fixed amount is divided randomly into two parts and the bigger part is assigned to one person, take the smaller part in the interval from \(0\) to half of the total amount.
Updated On: Jun 22, 2026
- \(\dfrac{2}{5}\)
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{2}{3}\)
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The Correct Option is D
Solution and Explanation
Step 1: Let the smaller part be given to \(Y\).
Since \(X\) is given the bigger part, let the amount received by \(Y\) be \(y\).
Then the amount received by \(X\) is
\[ 20000-y \]
Since \(Y\) gets the smaller part,
\[ 0\lt y\lt 10000 \]
Step 2: Apply the given condition.
We need the chance that \(X\) gets twice as much as \(Y\) or more.
So,
\[ X\geq 2Y \] \[ 20000-y\geq 2y \]
Step 3: Solve the inequality.
\[ 20000\geq 3y \] \[ y\leq \frac{20000}{3} \]
Step 4: Find the favorable interval.
The possible values of \(y\) lie in the interval
\[ (0,10000) \] The favorable values of \(y\) lie in the interval
\[ \left(0,\frac{20000}{3}\right] \]
Step 5: Calculate the required probability.
\[ P=\frac{\text{Length of favorable interval}}{\text{Length of total interval}} \] \[ P=\frac{\frac{20000}{3}}{10000} \] \[ P=\frac{2}{3} \]
Step 6: Final conclusion.
Hence, the required chance is
\[ \boxed{\frac{2}{3}} \]
Since \(X\) is given the bigger part, let the amount received by \(Y\) be \(y\).
Then the amount received by \(X\) is
\[ 20000-y \]
Since \(Y\) gets the smaller part,
\[ 0\lt y\lt 10000 \]
Step 2: Apply the given condition.
We need the chance that \(X\) gets twice as much as \(Y\) or more.
So,
\[ X\geq 2Y \] \[ 20000-y\geq 2y \]
Step 3: Solve the inequality.
\[ 20000\geq 3y \] \[ y\leq \frac{20000}{3} \]
Step 4: Find the favorable interval.
The possible values of \(y\) lie in the interval
\[ (0,10000) \] The favorable values of \(y\) lie in the interval
\[ \left(0,\frac{20000}{3}\right] \]
Step 5: Calculate the required probability.
\[ P=\frac{\text{Length of favorable interval}}{\text{Length of total interval}} \] \[ P=\frac{\frac{20000}{3}}{10000} \] \[ P=\frac{2}{3} \]
Step 6: Final conclusion.
Hence, the required chance is
\[ \boxed{\frac{2}{3}} \]
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