91Ó°ÊÓ

Here given that when a machine is adjusted properly then it produces 50% high quality and 50% medium quality items and usually, the machine is adjusted improperly 10% of the time and under this, it produces 25% high-quality items and 75% medium quality items.

Now five items are selected randomly at a certain time. Out of five four items are high quality and one is medium quality.

Then consider one more item which produces at the same time as the five items and it is also medium quality.

Let us consider,\({M_1}\)to be the event that the is properly adjusted at a certain time. And\({M_2}\)be the event that is improperly adjusted. Also consider\(H\)be the event that out of five items, four items are high-quality items.

So,

\(\Pr \left( {{M_1}} \right) = 0.9\)

\(\Pr \left( {{M_2}} \right) = 0.1\)

\(\Pr \left( {H|{M_1}} \right) = \left( {\begin{aligned}{{}{}}5\\4\end{aligned}} \right)\left( {{{0.5}^5}} \right)\)

\(\Pr \left( {H|{M_2}} \right) = \left( {\begin{aligned}{{}{}}5\\4\end{aligned}} \right)\left( {{{0.20}^4}} \right)\left( {0.75} \right)\)

a.

Now the probability that when we select five items randomly and out of them four are high quality and one is medium quality, the machine was in the properly adjusted mode is\(\Pr \left( {{M_1}|H} \right)\).

So,

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{|H}}} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}} \right){\bf{Pr}}\left( {{\bf{H|}}{{\bf{M}}_{\bf{1}}}} \right)}}{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}} \right){\bf{Pr}}\left( {{\bf{H|}}{{\bf{M}}_{\bf{1}}}} \right){\bf{ + Pr}}\left( {{{\bf{M}}_{\bf{2}}}} \right){\bf{Pr}}\left( {{\bf{H|}}{{\bf{M}}_{\bf{2}}}} \right)}}\\ = \frac{{\left( {0.9} \right) \times \left( {\begin{aligned}{{}{}}5\\4\end{aligned}} \right)\left( {{{0.5}^5}} \right)}}{{\left( {0.9} \right) \times \left( {\begin{aligned}{{}{}}5\\4\end{aligned}} \right)\left( {{{0.5}^5}} \right) + \left( {0.1} \right) \times \left( {\begin{aligned}{{}{}}5\\4\end{aligned}} \right)\left( {{{0.25}^4}} \right)\left( {0.75} \right)}}\\ = 0.989\end{aligned}\)

So, we can conclude that there is a 98% chance that at that certain time the machine was properly adjusted.

b.

Now consider another event G, that is a new item was selected randomly which was produced with the previously selected five items, found to be medium quality.

Therefore,\(\Pr \left( {G|{M_1}} \right) = 0.5\)

So, events\(H\)and\(G\)are conditionally independent of the events\({M_1}\)or\({M_2}\).

Now we have to find the conditional probability that for the both event\(H\)and\(G\), the machine was properly adjusted. That is,\(\Pr \left( {{M_1}|H,G} \right)\)

So,

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{|H,G}}} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{,G|H}}} \right)}}{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{,G|H}}} \right){\bf{ + Pr}}\left( {{{\bf{M}}_{\bf{2}}}{\bf{,G|H}}} \right)}}\\{\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{|H}}} \right){\bf{Pr}}\left( {{\bf{G|}}{{\bf{M}}_{\bf{1}}}{\bf{,H}}} \right)}}{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{|H}}} \right){\bf{Pr}}\left( {{\bf{G|}}{{\bf{M}}_{\bf{1}}}{\bf{,H}}} \right){\bf{ + Pr}}\left( {{{\bf{M}}_{\bf{2}}}{\bf{|H}}} \right){\bf{Pr}}\left( {{\bf{G|}}{{\bf{M}}_{\bf{2}}}{\bf{,H}}} \right)}}\\{\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{|H}}} \right){\bf{Pr}}\left( {{\bf{G|}}{{\bf{M}}_{\bf{1}}}} \right)}}{{{\bf{Pr}}\left( {{{\bf{M}}_{\bf{1}}}{\bf{|H}}} \right){\bf{Pr}}\left( {{\bf{G|}}{{\bf{M}}_{\bf{1}}}} \right){\bf{ + Pr}}\left( {{{\bf{M}}_{\bf{2}}}{\bf{|H}}} \right){\bf{Pr}}\left( {{\bf{G|}}{{\bf{M}}_{\bf{2}}}} \right)}}\end{aligned}\)

Therefore,

\(\begin{aligned}{}\Pr \left( {{M_1}|H,G} \right) = \frac{{0.989 \times 0.5}}{{0.989 \times 0.5 + \left( {1 - 0.989} \right) \times 0.75}}\\ = 0.9846\end{aligned}\)

Therefore, the required probability is 0.9846.