Probability
Independent Events
Grade 12
Question:
<p>Let two fair six-faced dice \(A\) and \(B\) be thrown simultaneously. If \(E_1\) is the event that die \(A\) shows up four, \(E_2\) is the event that die \(B\) shows up two and \(E_3\) is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT True?</p>
<p>\(E_2\) and \(E_3\) are independent</p>
<p>\(E_1\) and \(E_3\) are independent</p>
<p>\(E_1\), \(E_2\) and \(E_3\) are independent</p>
<p>\(E_1\) and \(E_2\) are independent</p>
Step-by-Step Solution
Key Concept: Test independence and mutual exclusivity of events by checking if P(A∩B) = P(A)·P(B) and whether events can occur simultaneously. E₁ and E₂ are independent, E₁ and E₃ are mutually exclusive (even sum when A=4), but E₂ and E₃ are independent.
<p><strong>Step 1: Define the events</strong></p><p>E₁: Die A shows 4 → P(E₁) = 1/6</p><p>E₂: Die B shows 2 → P(E₂) = 1/6</p><p>E₃: Sum is odd → P(E₃) = 18/36 = 1/2</p><p><strong>Step 2: Check E₁ and E₂</strong></p><p>E₁ ∩ E₂: A=4, B=2 (both can occur) → P(E₁ ∩ E₂) = 1/36</p><p>P(E₁)·P(E₂) = 1/6 × 1/6 = 1/36 ✓ Independent</p><p><strong>Step 3: Check E₁ and E₃</strong></p><p>If A=4 (even), sum = 4 + B is even for any B → Cannot be odd</p><p>P(E₁ ∩ E₃) = 0, but P(E₁)·P(E₃) = 1/12 ≠ 0 ✗ NOT independent</p><p>E₁ and E₃ are mutually exclusive</p><p><strong>Step 4: Check E₂ and E₃</strong></p><p>If B=2 (even), sum = A + 2 is odd when A is odd (1,3,5)</p><p>E₂ ∩ E₃: 3 outcomes → P(E₂ ∩ E₃) = 3/36 = 1/12</p><p>P(E₂)·P(E₃) = 1/6 × 1/2 = 1/12 ✓ Independent</p><p>∴ <strong>The statement that E₁ and E₃ are independent is NOT True (they are mutually exclusive)</strong></p>
Correct Answer: C