Three categories of candidates appear for an admission test: diligent (10%), lazy (30%) and confused (60%). A diligent candidate is 10 times more likely to clear the admission test compared to a lazy candidate.
If 40% of the candidates clearing the admission test are confused, what is the MAXIMUM possible value of the probability of a confused candidate clearing the test?
If 40% of the candidates clearing the admission test are confused, what is the MAXIMUM possible value of the probability of a confused candidate clearing the test?
Show Hint
- \( \frac{13}{37} \)
- \( \frac{6}{7} \)
- \( \frac{13}{90} \)
- \( \frac{37}{100} \)
- \( \frac{2}{3} \)
The Correct Option is A
Solution and Explanation
Let the total number of candidates be \( T \). Then: - The number of diligent candidates is \( 0.1T \), - The number of lazy candidates is \( 0.3T \), - The number of confused candidates is \( 0.6T \).
Step 2: Define the probabilities.
The probability of clearing the test for a diligent candidate is 10 times that of a lazy candidate. Let the probability of a lazy candidate clearing the test be \( p \), so the probability for a diligent candidate is \( 10p \), and the probability for a confused candidate is \( x \). We are given that 40% of the candidates who clear the test are confused, so: \[ \frac{0.6T \cdot x}{0.1T \cdot 10p + 0.3T \cdot p + 0.6T \cdot x} = 0.4 \] Simplifying the equation, we get: \[ \frac{0.6x}{1p + 3p + 6x} = 0.4 \]
Step 3: Solve for \( x \).
Solving the equation, we find: \[ x = \frac{13}{37} \]
Step 4: Conclusion.
The maximum possible value of the probability of a confused candidate clearing the test is \( \frac{13}{37} \). Therefore, the correct answer is (A).
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