Question

Three boxes contain lamp bulbs some of which are defective. The proportion of

defective in box A1; A2 and A3 are 1 2; 1 8 and 3 4 respectively. 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 A1 was selected?

Answer 1

Given that $A_1$ ,$A_2$ and $A_3$ represent the three boxes.

Then $P(A_1$ $)=P(A_2)=P(A_3)=$ $\frac{1}{30}$

Let Q be the event of drawing the defective bulb.

Then $P\left(\frac{Q}{A_1}\right)$$=P\left(drawing\:a\:defective\:bulb\:from\:A_1\right)=\frac{1}{2}$

$P\left(\frac{Q}{A_2}\right)=P\left(drawing\:a\:defective\:bulb\:from\:A_2\right)=\frac{1}{8}$

$P\left(\frac{Q}{A_3}\right)=P\left(drawing\:a\:defective\:bulb\:from\:A_3\right)=\frac{3}{4}$

The probability of selecting a defective bulb from $A_1$ being given that it is defective , is $P\left(\frac{A_1}{Q}\right).$

Then $P\left(\frac{A_1}{Q}\right)= \frac{P\left(A_1\right)\times P\left(\frac{Q}{A_1}\right)}{\left[P\left(A_1\right)\times P\left(\frac{Q}{A_1}\right)\right]+\left[P\left(A_2\right)\times \:P\left(\frac{Q}{A_2}\right)\right]+\left[P\left(A_3\right)\times \:P\left(\frac{Q}{A_3}\right)\right]}$

$P\left(\frac{A_1}{Q}\right)=$ $\frac{\frac{1}{3}\times \frac{1}{2}}{\left(\frac{1}{3}\times \:\frac{1}{2}\right)+\left(\frac{1}{3}\times \frac{1}{8}\right)+\left(\frac{1}{3}\times \frac{3}{4}\right)}$

$P\left(\frac{A_1}{Q}\right)= \frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{8}+\frac{3}{4}}=\frac{\frac{1}{2}}{\frac{4+1+6}{8}}=\frac{\frac{1}{2}}{\frac{11}{8}}$

$P\left(\frac{A_1}{Q}\right)=\frac{1}{2}\times \frac{8}{11}=\frac{4}{11}$ .