Question

Scrap rate of machine parts. A press produces parts used in the manufacture of large-screen plasma televisions. If the press is correctly adjusted, it produces parts with a scrap rate of 5%. If it is not adjusted correctly, it produces scrap at a 50% rate. From past company records, the machine is known to be correctly adjusted 90% of the time. A quality-control inspector randomly selects one part from those recently produced by the press and discovers it is defective. What is the probability that the machine is incorrectly adjusted?

Short answer

The probability that the machine is incorrectly adjusted is 0.53.

Step-by-step solution

Important formula

The formula for probability is$\mathbf{P}\mathbf{=}\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}$

The probability that the machine is incorrectly adjusted.

Here only components with a scrap rate of 5% or$\mathrm{P}\left(\mathrm{C}\right|\mathrm{M})=0.05$ are produced on the specified data, when the press is suitably calibrated. It produces scrap at a rate of 50% or $\mathrm{P}\left(\mathrm{C}\right|{\mathrm{M}}^{\mathrm{C}})=0.5$. If not calibrated correctly. The machine is known to be corrected 90% or$\mathrm{P}\left(\mathrm{M}\right)=0.9$ of the time from historical business registrations.

Now,

$\begin{array}{c}\mathrm{P}\left({\mathrm{M}}^{\mathrm{C}}\right)=1-\mathrm{P}\left(\mathrm{M}\right)\\ =1-0.9\\ =0.1\end{array}$

And

$\begin{array}{c}\mathrm{P}\left({\mathrm{M}}^{\mathrm{C}}\right|\mathrm{C})=\frac{\mathrm{P}\left({\mathrm{M}}^{\mathrm{C}}\right)\mathrm{P}\left(\mathrm{C}\right|{\mathrm{M}}^{\mathrm{C}})}{\mathrm{P}\left({\mathrm{M}}^{\mathrm{C}}\right)\mathrm{P}\left(\mathrm{C}\right|{\mathrm{M}}^{\mathrm{C}})+\mathrm{P}(\mathrm{M}\left)\mathrm{P}\right(\mathrm{C}\left|\mathrm{M}\right)}\\ =\frac{0.1+0.5}{(0.1)(0.5)+(0.9)(0.05)}\\ =0.53\end{array}$

Therefore, the probability that the machine is incorrectly adjusted is 0.53.