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?

Answer 1

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`