Probability

Let $E_1$ mean the bulb was made by unit $A$.Let $E_2$ mean the bulb was made by unit $B$.Let $E_3$ mean the bulb was made by unit $C$.Let $E$ mean the bulb is defective.The chance a bulb comes from each unit is: $P (E_1) = \frac{25}{100} (,A), P (E_2) = \frac{35}{100} (,B)$ from $P (E_3) = \frac{40}{100} (,C)$ from $P\left(\frac{E}{E_1}\right) = \frac{5}{100}, P\left(\frac{E}{E_2}\right) = \frac{4}{100}, P\left(\frac{E}{E_3}\right) = \frac{2}{100}.$, and $B$ from $P\left(\frac{E_2}{E}\right) = \frac{P(E_2) P\left(\frac{E}{E_2}\right)}{P(E_1) P\left(\frac{E}{E_1}\right) + P(E_2) P\left(\frac{E}{E_2}\right) + P(E_3) P\left(\frac{E}{E_3}\right)}$.The chance a bulb is defective from each unit is:$P\left(\frac{E_2}{E}\right) = \frac{\frac{35}{100} \times \frac{4}{100}}{\frac{25}{100} \times \frac{5}{100} + \frac{35}{100} \times \frac{4}{100} + \frac{40}{100} \times \frac{2}{100}}$To find how likely it is a defective bulb came from unit $35 \times 4 = 140$, use Bayes' theorem:$25 \times 5 = 125, 35 \times 4 = 140$ Now plug in the numbers:$40 \times 2 = 80$The top ( $\;P\left(\frac{E_2}{E}\right) = \frac{140}{125+140+80}$ ) and the bottom adds up ( $\;P\left(\frac{E_2}{E}\right) = \frac{140}{345} = \frac{28}{69}$, $\frac{28}{69}$ ) so:$\frac{28}{71}$$\frac{29}{67}$