AP EAMCET · Maths · Probability

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 $Rs . 20000$ in to two parts and gives $X$ the bigger part. Then, the chance that $X$ gets twice as much as $Y$ or more is

A $\frac{2}{5}$
B $\frac{1}{2}$
C $\frac{1}{3}$
D $\frac{2}{3}$

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Answer & Solution

Correct Answer

(D) $\frac{2}{3}$

Step-by-step Solution

Detailed explanation

Given, X gets the bigger part than Y So, the bigger part could be any number from 10000 to 20000. Now, if the bigger part is to be at least twice as much as the smaller part, we have X≥2Y ⇒X≥220000-X ⇒X≥400003 So, the probability is equal to…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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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 in to two parts and gives X the bigger part. Then, the chance that X gets twice as much as Y or more is

Step-by-step Solution