The chances of a defective screw in three boxes A, B, and C are 1/5 , 1/6 , and 1/7 respectively. A box is selected at random and a screw is found to - Mathematics MCQ Mathematics Mock Test - 1 2026 | ExamTest.live

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The chances of a defective screw in three boxes A, B, and C are 1/5, 1/6, and 1/7 respectively. A box is selected at random and a screw is found to be defective. Find the probability that it came from box A.

Solution & Step-by-step Explanation

Let E_{1}, E_{2}, E_{3} be the events of selecting box A, B, and C. Since selection is at random, P (E_{1}) = P (E_{2}) = P (E_{3}) = 1/3 .Let D be the event that the screw is defective. P (D ∣ E_{1}) = 1/5, P (D ∣ E_{2}) = 1/6, P (D ∣ E_{3}) = 1/7 .Using Bayes' Theorem: P (E_{1} ∣ D) = \frac{P ( E _{1} ) P ( D ∣ E _{1} )}{P ( E _{1} ) P ( D ∣ E _{1} ) + P ( E _{2} ) P ( D ∣ E _{2} ) + P ( E _{3} ) P ( D ∣ E _{3} )} P (E_{1} ∣ D) = \frac{\frac{1}{3} \cdot \frac{1}{5}}{\frac{1}{3} \cdot \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{7}} Cancelling 1/3 : P (E_{1} ∣ D) = \frac{1/5}{1/5 + 1/6 + 1/7} = \frac{1/5}{\frac{42 + 35 + 30}{210}} = \frac{1/5}{107/210} P (E_{1} ∣ D) = \frac{1}{5} \cdot \frac{210}{107} = \frac{42}{107}

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The chances of a defective screw in three boxes A, B, and C are 1/5, 1/6, and 1/7 respectively. A box is selected at random and a screw is found to be defective. Find the probability that it came from box A.

40/107

41/107

42/107

43/107

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