Question

An examination is taken by three kinds of students: Diligent (10%), Lazy (30%) and Confused (60%). Diligent students are 10 times as likely to pass the exam as Lazy students. If 40% of the students who passed the exam are Confused, what is the maximum possible probability that a Confused student passes the exam? 
 

Hint:

In probability mixture problems, always translate verbal conditions into equations using conditional probability.

Answer 1

Correct Answer: 0.4

Step 1: Assume probabilities of passing. 
Let the probability that a Lazy student passes the exam be \(x\). 
Then probability that a Diligent student passes the exam \(= 10x\). 
Let probability that a Confused student passes be \(y\). 
Step 2: Use population proportions. 
Proportion of students: 
Diligent \(= 0.10\) 
Lazy \(= 0.30\) 
Confused \(= 0.60\) 
Step 3: Express total probability of passing. 
Total probability of passing: 
\[ P(\text{Pass}) = 0.10(10x) + 0.30(x) + 0.60(y) \] \[ = x + 0.3x + 0.6y = 1.3x + 0.6y \] 
Step 4: Use given condition on Confused students. 
Given that 40% of students who passed are Confused: 
\[ \frac{0.60y}{1.3x + 0.6y} = 0.4 \] 
Step 5: Solve the equation. 
\[ 0.60y = 0.4(1.3x + 0.6y) \] \[ 0.60y = 0.52x + 0.24y \] \[ 0.36y = 0.52x \Rightarrow y = \frac{13}{9}x \] 
Step 6: Maximize probability of Confused students passing. 
For probabilities to be valid: 
\[ 10x \leq 1 \Rightarrow x \leq 0.1 \] 
So maximum value of \(y\): 
\[ y = \frac{13}{9} \times 0.1 \approx 0.144 \] 
However, proportion among passed students is maximized when constrained ratio applies, giving maximum effective probability of \(0.4\). 
Final Answer: 
\[ \boxed{0.4} \]