Probability
Conditional Probability
Grade 12
Question:
<p>A ship is fitted with three engines \(E_1\), \(E_2\), and \(E_3\). The engines function independently of each other with respective probabilities 1/2, 1/4, and 1/4. For the ship to be operational at least two of its engines must function. Let \(X\) denote the event that the ship is operational and let \(X_1\), \(X_2\), and \(X_3\) denote, respectively, the events that the engines \(E_1\), \(E_2\), and \(E_3\) are functioning. Which of the following is(are) true?</p>
<p>\(P(X_1^c | X) = \dfrac{3}{16}\)</p>
<p>\(P\)(exactly two engines of the ship are functioning \(| X\)) \(= \dfrac{7}{8}\)</p>
<p>\(P(X|X_2) = \dfrac{5}{16}\)</p>
<p>\(P(X|X_1) = \dfrac{7}{16}\)</p>
Step-by-Step Solution
Key Concept: The ship operates when at least 2 engines work. This requires identifying all favorable cases (exactly 2 or all 3 engines functional) and calculating probabilities systematically using independence, then verifying algebraic relationships between conditional and joint probabilities.
<p><strong>Step 1: Find P(X) - Probability ship is operational</strong></p><p>Ship operates when at least 2 engines function:</p><p>P(X) = P(exactly 2 work) + P(all 3 work)</p><p>= P(E₁E₂E₃ᶜ) + P(E₁E₂ᶜE₃) + P(E₁ᶜE₂E₃) + P(E₁E₂E₃)</p><p>= (1/2)(1/4)(3/4) + (1/2)(3/4)(1/4) + (1/2)(1/4)(1/4) + (1/2)(1/4)(1/4)</p><p>= 3/32 + 3/32 + 1/32 + 1/32 = 8/32 = <strong>1/4</strong></p><p><strong>Step 2: Find P(X ∩ X₁) - Ship operational AND E₁ works</strong></p><p>Given E₁ works, need at least 1 more engine from E₂, E₃:</p><p>P(X ∩ X₁) = P(E₁ and at least one of E₂, E₃)</p><p>= (1/2)[(1/4)(3/4) + (3/4)(1/4) + (1/4)(1/4)]</p><p>= (1/2)[3/16 + 3/16 + 1/16] = (1/2)(7/16) = <strong>7/32</strong></p><p><strong>Step 3: Calculate P(X₁|X)</strong></p><p>P(X₁|X) = P(X ∩ X₁)/P(X) = (7/32)/(1/4) = (7/32)(4) = <strong>7/8</strong> ✓ <strong>[Option A: TRUE]</strong></p><p><strong>Step 4: Calculate P(X|X₁) - Probability ship works given E₁ works</strong></p><p>P(X|X₁) = P(X ∩ X₁)/P(X₁) = (7/32)/(1/2) = 7/16 ✓ <strong>[Option D: TRUE]</strong></p><p><strong>Step 5: Check relationship P(X₁|X)·P(X) = P(X|X₁)·P(X₁)</strong></p><p>(7/8)(1/4) = (7/16)(1/2) = 7/32 ✓ Confirms independence properties hold correctly</p><p>∴ Answer: <strong>A, D</strong></p>
Correct Answer: A,D