<p>Two numbers are randomly selected and multiplied. Consider two events \(E_1\) and \(E_2\) defined as<br>\(E_1\): Their product is divisible by 5<br>\(E_2\): Unit's places in their product is 5<br>Which of the following statement is/are correct?</p>
<p>\(E_1\) is twice as likely to occur as \(E_2\)</p>
<p>\(E_1\) and \(E_2\) are disjoint</p>
<p>\(P(E_2/E_1) = 1/4\)</p>
<p>\(P(E_1/E_2) = 1\)</p>
Step-by-Step Solution
Key Concept: Event E₂ (unit's digit is 5) is a subset of E₁ (divisible by 5), since a number with unit's digit 5 is always divisible by 5. However, E₁ includes cases where the product ends in 0 (divisible by 5 but unit's digit ≠ 5), making E₁ ⊃ E₂ with E₂ ≠ E₁.
<p><strong>Step 1:</strong> Analyze E₁ (product divisible by 5)</p><p>A number is divisible by 5 if its unit's digit is 0 or 5. For a product to be divisible by 5, at least one factor must be divisible by 5 (i.e., end in 0 or 5).</p><p><strong>Step 2:</strong> Analyze E₂ (product's unit's digit is 5)</p><p>For the product to have unit's digit 5, both numbers must be odd and at least one must end in 5. Examples: 3×5=15, 5×7=35, 5×15=75. (Note: even × anything = even unit's digit; two odds ending in 5: 5×5=25)</p><p><strong>Step 3:</strong> Determine relationship between E₁ and E₂</p><p>If E₂ occurs (unit's digit is 5), then E₁ must occur (divisible by 5). Therefore E₂ ⊂ E₁.</p><p>However, if product ends in 0 (like 2×5=10, 4×5=20), then E₁ occurs but E₂ does not.</p><p>Thus: E₂ ⊂ E₁ and E₂ ≠ E₁, meaning E₁ and E₂ are NOT the same event.</p><p><strong>Step 4:</strong> Evaluate statements</p><p>• E₂ ⊂ E₁ is TRUE (every product with unit's digit 5 is divisible by 5)</p><p>• P(E₂) < P(E₁) is TRUE (E₂ is a proper subset)</p><p>• E₁ and E₂ are not mutually exclusive (they can occur together)</p><p>∴ Correct statements identify the subset relationship and probability inequality.</p>
Correct Answer: D