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Question
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 `1/2`, `1/8` and `3/4`. A box is selected at random and a bulb drawn from it. If the selected bulb is found to be defective, what is the probability that box B1 was selected?
Solution
Given that B1, B2 and B3 represent three boxes.
Then P(B1) = P(B2) = P(B3) = `1/3`
Let A be the event of selecting a defective bulb.
Then P`("A"/"B"_1)` = P(drawing a defective bulb from B1) = `1/2`
P`("A"/"B"_2)` = P(drawing a defective bulb from B2) = `1/8`
P`("A"/"B"_3)` = P(drawing a defective bulb from B3) = `3/4`
The probability of drawing a defective bulb from B1, being given that it is defective, is `"P"("B"_1/"A")`.
`"P"("B"_1/"A") = ("P"("B"_1) xx "P"("A"/"B"_1))/("P"("B"_1) xx "P"("A"/("B"_1)) + "P"("B"_2) xx "P"("A"/("B"_2)) + "P"("B"_3) xx "P"("A"/("B"_3))`
= `(1/3 xx 1/2)/(1/3 xx 1/2 + 1/3 xx 1/8 + 1/3 xx 3/4)`
= `(1/2)/(1/2 + 1/8 + 3/4)`
= `(1/2)/(((4 + 1 + 6)/8))`
= `(1/2)/(11/8)`
= `1/2 xx 8/11`
= `4/11`
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