Question

If a machine is correctly set up, it produces $90\%$ acceptable items. If it is incorrectly set up, it produces only $40\%$ acceptable items. Past experience shows that $80\%$ of the set ups are correctly done. If after a certain set up, the machine produces $2$ acceptable items, find the probability that the machine is correctly setup.

Answer 1

Let $E_1$ = Event that the reactions is correctly setup.

$E_2$ = Event that the machine is incorrectly setup.

$E$ = Event that machine produce two acceptable items.

Now,

$P\left(E_1/E\right)$ = Probability of machine have a correct setup if it produce two acceptable item.

$=\frac{P\left(E_1\right)\cdot P\left(E/E_1\right)}{P\left(E_1\right)\cdot P\left(E/E_1\right)+P\left(E_2\right)\cdot P\left(E/E_2\right)}$

$P\left(E_1\right)=80\%=0.8$

$P\left(E_2\right)=\left(100-80\right)=20\%=0.2$

$P\left(E/E_1\right)=90\%\times 90\%=9\times 9=0.81$

$P\left(E/E_2\right)=40\%\times 40\%=0.4\times 0.4=0.16$

$\therefore P\left(E_1/E\right)=\frac{0.8\times 0.81}{0.8\times 0.81+0.2\times 0.16}=0.95$