Question

In a class, 60% of the students know lesson I, 40% know lesson II and 20% know lesson I and lesson II. A student is selected at random. The probability that the student does not know lesson I and lesson II is:

• A. \( 0 \)
• B. \( \frac{4}{5} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{1}{5} \)
• E. \( \frac{2}{5} \)

Hint:

Drawing a Venn diagram is the most intuitive way to visualize these problems. Place the intersection value (20%) first, then fill the remaining sections (40% for 'Only I' and 20% for 'Only II'). The total inside the circles is 80%, leaving 20% outside.

Answer 1

The Correct Option is D

Concept: This problem can be solved using the Set Theory principle of inclusion-exclusion for probability. Let \( P(A) \) be the probability of a student knowing lesson I and \( P(B) \) be the probability of a student knowing lesson II. The probability of knowing at least one lesson is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). The probability of knowing neither is the complement: \( 1 - P(A \cup B) \).

Step 1: Calculate the probability of knowing at least one lesson.
Given:
• \( P(A) = 60\% = 0.6 \)
• \( P(B) = 40\% = 0.4 \)
• \( P(A \cap B) = 20\% = 0.2 \) Applying the formula: \[ P(A \cup B) = 0.6 + 0.4 - 0.2 = 0.8 \] This means 80% of the students know either lesson I, lesson II, or both.

Step 2: Find the probability of knowing neither lesson.
The question asks for the probability that a student does not know lesson I and does not know lesson II, which is \( P(A' \cap B') \). By De Morgan's Law, this is equal to \( P(A \cup B)' \): \[ P(\text{Neither}) = 1 - P(A \cup B) \] \[ P = 1 - 0.8 = 0.2 \] Converting the decimal back to a fraction: \[ 0.2 = \frac{2}{10} = \frac{1}{5} \]