91Ó°ÊÓ

The number of different machines used for producing a large batch of similar manufactured products is 3, labelled by\({M_1},{M_2},{M_3}\).

20% of the items were produced by machine 1, 30% of the items were produced by machine 2, and 50% of the products were produced by machine 3.

And also said that 1% of items are defectively made by machine 1, 2% of the products are defectively made by machine 2 and 3% of products are defectively made by machine 3.

One item was selected at random from a batch of manufactured items and found to be defective.

As given in the example 2.3.4, we know that\({B_i}\)is the event of selecting an item, produced by machine\({M_i}\;Where\;i = \left( {1,2,3} \right)\). And A be the event of selecting a defective item.

So, we know,

\(P\left( {{B_1}} \right) = 0.20\),\(P\left( {{B_2}} \right) = 0.30\),\(P\left( {{B_3}} \right) = 0.50\)

And also know that the probability of an item produced by machine\({M_i}\), will be defective. That is,

\(P\left( {A|{B_1}} \right) = 0.01\),\(P\left( {A|{B_2}} \right) = 0.02\),\(P\left( {A|{B_3}} \right) = 0.03\)

And the probability that a randomly chosen defective item is produced by machine 2 is, \(P\left( {{B_2}|A} \right) = 0.26\)

Now we have to calculate theprobabilities that the randomly selected item is manufacturedby machine 1 and machine 2. And the posterior probability for machine 2 is already given that is 0.26

So,

\(\begin{aligned}{}P\left( {{B_1}|A} \right) &= \frac{{P\left( {{B_1}} \right)P\left( {A|{B_2}} \right)}}{{\sum\limits_{i = 1}^3 {P\left( {{B_i}} \right)P\left( {A|{B_i}} \right)} }}\\ &= \frac{{0.2 \times 0.01}}{{\left( {0.2 \times 0.01} \right) + \left( {0.3 \times 0.02} \right) + \left( {0.5 \times 0.03} \right)}}\\ &= 0.086\end{aligned}\)

And,

\(\begin{aligned}{}P\left( {{B_3}|A} \right) &= \frac{{P\left( {{B_3}} \right)P\left( {A|{B_3}} \right)}}{{\sum\limits_{i = 1}^3 {P\left( {{B_i}} \right)P\left( {A|{B_i}} \right)} }}\\ &= \frac{{0.5 \times 0.03}}{{\left( {0.2 \times 0.01} \right) + \left( {0.3 \times 0.02} \right) + \left( {0.5 \times 0.03} \right)}}\\ &= 0.65\end{aligned}\)

So, we can see that the probability of the defected item is from machine 2 is 0.08 and from machine 3 that is 0.65.

In the probability theory the prior probabilities are the probabilities to know prior to obtaining some additional data or information. And posterior probabilities are the probabilities of events after obtaining the additional data.

Here\({\bf{P}}\left( {{{\bf{B}}_{\bf{1}}}} \right){\bf{,P}}\left( {{{\bf{B}}_{\bf{2}}}} \right)\;{\bf{and}}\;{\bf{P}}\left( {{{\bf{B}}_{\bf{3}}}} \right)\)are the prior probabilities and\({\bf{P}}\left( {{{\bf{B}}_{\bf{1}}}{\bf{|A}}} \right){\bf{,P}}\left( {{{\bf{B}}_{\bf{2}}}{\bf{|A}}} \right)\;{\bf{and}}\;{\bf{P}}\left( {{{\bf{B}}_{\bf{3}}}{\bf{|A}}} \right)\)are the posterior probabilities.

We can clearly see that the posterior probability for machine 3, that is \(\left( {i = 3} \right)\)is larger than its prior probability that is \(P\left( {{B_3}|A} \right) > P\left( {{B_3}} \right)\). As machine 3 produced a large proportion of items so the probability of getting a defective item also increases for machine 3.