Question:
A student selected at random did not qualify the examination. Find the probability that the student was a regular student.
A student selected at random did not qualify the examination. Find the probability that the student was a regular student.
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Check if the denominators of parts (iii)(a) and (iii)(b) sum to 1. Here \( 0.08 + 0.92 = 1.0 \), which confirms total probability consistency.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Total probability of the complementary event.
Step 1: Identify probabilities of not qualifying
Let \( \bar{A} \) be the event of not qualifying. From dropouts: \( P(\bar{A}|E_1) = 0.95 \). From regular students: \( P(\bar{A}|E_2) = 1 - 0.10 = 0.90 \).
Step 2: Calculate total probability of not qualifying
\[ P(\bar{A}) = P(E_1)P(\bar{A}|E_1) + P(E_2)P(\bar{A}|E_2) \] \[ P(\bar{A}) = (0.40 \times 0.95) + (0.60 \times 0.90) \] \[ P(\bar{A}) = 0.38 + 0.54 = 0.92 \]
Step 3: Apply Bayes' Theorem
We need to find \( P(E_2|\bar{A}) \): \[ P(E_2|\bar{A}) = \frac{P(E_2)P(\bar{A}|E_2)}{P(\bar{A})} \] \[ P(E_2|\bar{A}) = \frac{0.60 \times 0.90}{0.92} = \frac{0.54}{0.92} \] \[ P(E_2|\bar{A}) = \frac{54}{92} = \frac{27}{46} \approx 0.587 \] The probability is 27/46.
• Total probability of the complementary event.
Step 1: Identify probabilities of not qualifying
Let \( \bar{A} \) be the event of not qualifying. From dropouts: \( P(\bar{A}|E_1) = 0.95 \). From regular students: \( P(\bar{A}|E_2) = 1 - 0.10 = 0.90 \).
Step 2: Calculate total probability of not qualifying
\[ P(\bar{A}) = P(E_1)P(\bar{A}|E_1) + P(E_2)P(\bar{A}|E_2) \] \[ P(\bar{A}) = (0.40 \times 0.95) + (0.60 \times 0.90) \] \[ P(\bar{A}) = 0.38 + 0.54 = 0.92 \]
Step 3: Apply Bayes' Theorem
We need to find \( P(E_2|\bar{A}) \): \[ P(E_2|\bar{A}) = \frac{P(E_2)P(\bar{A}|E_2)}{P(\bar{A})} \] \[ P(E_2|\bar{A}) = \frac{0.60 \times 0.90}{0.92} = \frac{0.54}{0.92} \] \[ P(E_2|\bar{A}) = \frac{54}{92} = \frac{27}{46} \approx 0.587 \] The probability is 27/46.
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