JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Probability

In a bolt factory, machines A, B and C manufacture respectively

20%

30%

and

50%

of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine C is

(A) $\frac{2}{7}$
(B) $\frac{9}{28}$
(C) $\frac{5}{14}$ ✓ Correct
(D) $\frac{3}{7}$

Answer: (C)

Solution

Using Bayes' theorem:

P (C ∣ D) = \frac{P ( D ∣ C ) \cdot P ( C )}{P ( D )}

. We have

P (A) = 0.2, P (B) = 0.3, P (C) = 0.5

and

P (D ∣ A) = 0.03, P (D ∣ B) = 0.04, P (D ∣ C) = 0.02

. By the law of total probability:

P (D) = P (D ∣ A) P (A) + P (D ∣ B) P (B) + P (D ∣ C) P (C) = 0.03 \times 0.2 + 0.04 \times 0.3 + 0.02 \times 0.5 = 0.006 + 0.012 + 0.010 = 0.028

. Therefore

P (C ∣ D) = \frac{0.02 \times 0.5}{0.028} = \frac{0.01}{0.028} = \frac{10}{28} = \frac{5}{14}

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Probability

Solution

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