In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and…
2021
In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?
- A.
0.55
- B.
0.45
- C.
0.31
- D.
0.41
Attempted by 7 students.
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Correct answer: D
To solve this problem, we use Bayes' Theorem, which allows us to find the conditional probability of an event based on prior knowledge of conditions related to the event.
Step-by-Step Calculation
Identify the probabilities of manufacture:
P(P) = 0.25
P(Q) = 0.35
P(R) = 0.40
Identify the probabilities of a defect given the machine (conditional probability):
P(Defect|P) = 0.05
P(Defect|Q) = 0.04
P(Defect|R) = 0.02
Calculate the total probability of drawing a defective bulb :
P(Defect) = P(P) * P(Defect|P) + P(Q) * P(Defect|Q) + P(R) * P(Defect|R)
P(Defect) = (0.25 * 0.05) + (0.35 * 0.04) + (0.40 * 0.02)
P(Defect) = 0.0125 + 0.0140 + 0.0080
P(Defect) = 0.0345
Apply Bayes' Theorem to find P(Q|Defect):
P(Q|Defect) = [P(Q) * P(Defect|Q)] / P(Defect)
P(Q|Defect) = (0.35 * 0.04) / 0.0345
P(Q|Defect) = 0.0140 / 0.0345
P(Q|Defect) ≈ 0.4057, which rounds to 0.41.