Question:
A student selected at random qualified the examination. Find the probability that student is not a dropout.
A student selected at random qualified the examination. Find the probability that student is not a dropout.
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Bayes' Theorem is essentially (Target Path Probability) (Sum of all Path Probabilities).
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Bayes' Theorem: \( P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum P(E_j)P(A|E_j)} \).
Step 1: List the known probabilities
Probability of being a dropout: \( P(E_1) = 0.40 \). Probability of being a regular student: \( P(E_2) = 0.60 \). Probability of qualifying given dropout: \( P(A|E_1) = 0.05 \). Probability of qualifying given regular: \( P(A|E_2) = 0.10 \).
Step 2: Calculate total probability of qualifying
\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \] \[ P(A) = (0.40 \times 0.05) + (0.60 \times 0.10) \] \[ P(A) = 0.02 + 0.06 = 0.08 \]
Step 3: Apply Bayes' Theorem to find \( P(\text{Regular}|\text{Qualified}) \)
The phrase "not a dropout" means the student is a regular student (\( E_2 \)). \[ P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(A)} \] \[ P(E_2|A) = \frac{0.60 \times 0.10}{0.08} = \frac{0.06}{0.08} \] \[ P(E_2|A) = \frac{6}{8} = 0.75 \] The probability is 0.75.
• Bayes' Theorem: \( P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum P(E_j)P(A|E_j)} \).
Step 1: List the known probabilities
Probability of being a dropout: \( P(E_1) = 0.40 \). Probability of being a regular student: \( P(E_2) = 0.60 \). Probability of qualifying given dropout: \( P(A|E_1) = 0.05 \). Probability of qualifying given regular: \( P(A|E_2) = 0.10 \).
Step 2: Calculate total probability of qualifying
\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \] \[ P(A) = (0.40 \times 0.05) + (0.60 \times 0.10) \] \[ P(A) = 0.02 + 0.06 = 0.08 \]
Step 3: Apply Bayes' Theorem to find \( P(\text{Regular}|\text{Qualified}) \)
The phrase "not a dropout" means the student is a regular student (\( E_2 \)). \[ P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(A)} \] \[ P(E_2|A) = \frac{0.60 \times 0.10}{0.08} = \frac{0.06}{0.08} \] \[ P(E_2|A) = \frac{6}{8} = 0.75 \] The probability is 0.75.
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