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(\(\frac{B_1}{A}\))