Probability
Bayes' Theorem
Grade 12
Question:
<p>A computer producing factory has only two plants \(T_1\) and \(T_2\). Plant \(T_1\) produces 20% and plant \(T_2\) produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that \(P\) (computer turns out to be defective given that it is produced in plant \(T_1\)) = 10\(P\) (computer turns out to be defective given that it is produced in plant \(T_2\)), where \(P(E)\) denotes the probability of an event \(E\). A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant \(T_2\) is</p>
<p>\(\dfrac{36}{73}\)</p>
<p>\(\dfrac{47}{79}\)</p>
<p>\(\dfrac{78}{96}\)</p>
<p>\(\dfrac{75}{83}\)</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem with the law of total probability. First, find the individual defect rates from the constraint P(D|T₁) = 10·P(D|T₂), then apply Bayes to find P(T₂|D') where D' is non-defective.
<p><strong>Step 1:</strong> Set up variables. Let P(D|T₁) = p₁ and P(D|T₂) = p₂. Given: p₁ = 10p₂</p><p><strong>Step 2:</strong> Use the law of total probability for defective computers:</p><p>P(D) = P(D|T₁)·P(T₁) + P(D|T₂)·P(T₂)</p><p>0.07 = p₁(0.2) + p₂(0.8)</p><p>0.07 = 10p₂(0.2) + p₂(0.8) = 2p₂ + 0.8p₂ = 2.8p₂</p><p>∴ p₂ = 0.07/2.8 = 1/40 = 0.025</p><p>∴ p₁ = 10(0.025) = 0.25</p><p><strong>Step 3:</strong> Find P(D') = 1 - P(D) = 1 - 0.07 = 0.93</p><p><strong>Step 4:</strong> Find P(D'|T₁) = 1 - 0.25 = 0.75 and P(D'|T₂) = 1 - 0.025 = 0.975</p><p><strong>Step 5:</strong> Apply Bayes' theorem for non-defective computers:</p><p>P(T₂|D') = P(D'|T₂)·P(T₂) / P(D')</p><p>P(T₂|D') = (0.975)(0.8) / 0.93 = 0.78 / 0.93 = 78/93 = 26/31</p><p>∴ Answer: D</p>
Correct Answer: D