Question

Suppose that in Example 2.3.4 in this section, the item selected at random from the entire lot is found to be non-defective. Determine the posterior probability that it was produced by machine\({{\bf{M}}_{\bf{2}}}\).

Short answer

The posterior probability that the randomly selected item was produced by the machine \({M_2}\) is 0.301.

Step-by-step solution

Given information

Refer to the example 2.3.4, probabilities of selecting an item from machine\({M_i};\;{\rm{for}}\;i = 1,2,3\)are;

\(\begin{aligned}{}\Pr \left( {{B_1}} \right) = 0.2\\\Pr \left( {{B_2}} \right) = 0.3\\\Pr \left( {{B_3}} \right) = 0.5\end{aligned}\)

If A as the event of selecting defective item, then,

\(\begin{aligned}{}\Pr \left( {A\left| {{B_1}} \right.} \right) = 0.01\\\Pr \left( {A\left| {{B_2}} \right.} \right) = 0.02\\\Pr \left( {A\left| {{B_3}} \right.} \right) = 0.03\end{aligned}\)

The randomly selected item is non-defective.

State the required event

The posterior probability that it was produced by machine \({M_2}\) is expressed as,\(\Pr \left( {{B_2}\left| {{A^c}} \right.} \right)\) , where \({A^c}\) is that the selected item is non-defective.

Compute the probability

The probability of a non-defective item being produced by machine 1,2,3 respectively are,

\(\begin{aligned}{}{\bf{P}}\left( {{{\bf{A}}^{\bf{c}}}\left| {{{\bf{B}}_{\bf{1}}}} \right.} \right){\bf{ = 1 - P}}\left( {{\bf{A}}\left| {{{\bf{B}}_{\bf{1}}}} \right.} \right)\\ = 1 - 0.01\\ = 0.99\end{aligned}\)

\(\begin{aligned}{}{\bf{P}}\left( {{{\bf{A}}^{\bf{c}}}\left| {{{\bf{B}}_2}} \right.} \right){\bf{ = 1 - P}}\left( {{\bf{A}}\left| {{{\bf{B}}_2}} \right.} \right)\\ = 1 - 0.02\\ = 0.98\end{aligned}\)

\(\begin{aligned}{}{\bf{P}}\left( {{{\bf{A}}^{\bf{c}}}\left| {{{\bf{B}}_3}} \right.} \right){\bf{ = 1 - P}}\left( {{\bf{A}}\left| {{{\bf{B}}_3}} \right.} \right)\\ = 1 - 0.03\\ = 0.97\end{aligned}\)

Using Bayes’ theorem, the probability is that the randomly selected non-defective item is produced by machine 2 is,

\(\begin{aligned}{}{\bf{P}}\left( {{{\bf{B}}_{\bf{2}}}\left| {{{\bf{A}}^{\bf{c}}}} \right.} \right){\bf{ = }}\frac{{{\bf{P}}\left( {{{\bf{B}}_{\bf{2}}}} \right){\bf{P}}\left( {{{\bf{A}}^{\bf{c}}}{\bf{|}}{{\bf{B}}_{\bf{2}}}} \right)}}{{{\bf{P}}\left( {{{\bf{B}}_{\bf{1}}}} \right){\bf{P}}\left( {{{\bf{A}}^{\bf{c}}}{\bf{|}}{{\bf{B}}_{\bf{1}}}} \right){\bf{ + P}}\left( {{{\bf{B}}_{\bf{2}}}} \right){\bf{P}}\left( {{{\bf{A}}^{\bf{c}}}{\bf{|}}{{\bf{B}}_{\bf{2}}}} \right){\bf{ + P}}\left( {{{\bf{B}}_{\bf{3}}}} \right){\bf{P}}\left( {{{\bf{A}}^{\bf{c}}}{\bf{|}}{{\bf{B}}_{\bf{3}}}} \right)}}\\ = \frac{{\left( {0.3 \times 0.98} \right)}}{{\left( {0.2 \times 0.99} \right) + \left( {0.3 \times 0.98} \right) + \left( {0.5 \times 0.97} \right)}}\\ = 0.301\end{aligned}\)

Thus, there is approximately a 30.1% of chance that the selected non-defective item is from machine 2.