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in Descriptive Statistics and Probability by (47.4k points)
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Three boxes B1, B2, B3 contain lamp bulbs some of which are defective. The defective proportions in box B1, box B2 and box B3 are respectively \(\frac{1}{2},\frac{1}{3}\), and \(\frac{1}{4}\). A box is selected at random and a bulb is drawn from it. If the selected bulb is found to be defective, what is the probability that box B1 was selected?

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by (47.8k points)
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Best answer

Given that B1, B2 and B3 represent three boxes. 

Then P(B1) = P(B2) = P(B3) = \(\frac{1}{3}\)

Let A be the event of selecting a defective bulb. 

Then P(\(\frac{A}{B_1}\)) = P(drawing a defective bulb from B1) = \(\frac{1}{2}\)

P(\(\frac{A}{B_2}\)) = P(drawing a defective bulb from B2) = \(\frac{1}{8}\)

P(\(\frac{A}{B_3}\)) = P(drawing a defective bulb from B3) = \(\frac{3}{4}\)

The probability of drawing a defective bulb from B1, being given that it is defective, is P(\(\frac{B_1}{A}\)).

 P(\(\frac{B_1}{A}\)

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