Mathematics - Area Under Curves Question with Solution | TestHub

Let $E_1,E_2$ and $E_3$ denote the events of selecting boxes $A,B,C$ respectively and $A$ be the event that a screw selected at random is defective. Then, $$\begin{array}{l}P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{1}{3},P\left(E_3\right)=\frac{1}{3}\\ P\left(\frac{A}{E_1}\right)=\frac{1}{5},P\left(\frac{A}{E_2}\right)=\frac{1}{6},P\left(\frac{A}{E_3}\right)=\frac{1}{7}\end{array}$$ By Baye's rule, the required probability $$\begin{array}{rl}P\left(\frac{E_1}{A}\right)=& \frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_3\right)P\left(\frac{A}{E_3}\right)}\\ \Rightarrow P\left(\frac{E_1}{A}\right)& =\frac{\frac{1}{3}\cdot \frac{1}{5}}{\frac{1}{3}\cdot \frac{1}{5}+\frac{1}{3}\cdot \frac{1}{6}+\frac{1}{3}\cdot \frac{1}{7}}\\ & =\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}=\frac{42}{107}\end{array}$$