91Ó°ÊÓ

Let \(\theta \) the proportion of defective items in a large manufactured lot is either 0.1 or 0.2

Let 8 items be selected at random from the lot. From which exactly two of them are defective.

Then from the definition of proportion of defective items.\({X_1}...{X_n}\)form n Bernoulli trails with parameter\(\theta \)

The p.f. of the each observations\({X_i}\) is given

\(f\left( {x|\theta } \right) = \left\{ \begin{aligned}{l}{\theta ^x}{\left( {1 - \theta } \right)^{1 - x}}\;\;\;\;\;{\rm{for}}\;x = 0,1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}.\end{aligned} \right.\)

If\(y = \sum\limits_{i = 1}^n {{x_I}} \)then the joint pf of\({X_1},...{X_n}\)can be written in the form for\({x_i} = 0\;{\rm{or}}\;1\):

\({f_n}\left( {x|\theta } \right) = {\theta ^y}{\left( {1 - \theta } \right)^{n - y}}\)

Since n = 8 and y = 2

\(\begin{aligned}{f_n}\left( {x|\theta } \right) = {\theta ^2}{\left( {1 - \theta } \right)^{8 - 2}}\\ = {f_n}\left( {x|\theta } \right) = {\theta ^2}{\left( {1 - \theta } \right)^6}\end{aligned}\)

Definition of the posterior pf is:

\(\)\(\xi \left( {\theta |x} \right) = \frac{{f\left( {{x_1}|\theta } \right)...f\left( {{x_n}|\theta } \right)\xi \left( \theta \right)}}{{{g_n}\left( x \right)}}\)

Therefore,

\(\begin{aligned}\xi \left( {0.1|x} \right) = {\rm P}\left( {\theta = 0.1|x} \right)\\ = \frac{{\xi \left( {0.1} \right){f_n}\left( {x|0.1} \right)}}{{\xi \left( {0.1} \right){f_n}\left( {x|0.1} \right) + \xi \left( {0.2} \right){f_n}\left( {x|0.2} \right)}}\\ = \frac{{\left( {0.7} \right){{\left( {0.1} \right)}^2}{{\left( {0.9} \right)}^6}}}{{\left( {0.7} \right){{\left( {0.1} \right)}^2}{{\left( {0.9} \right)}^6} + \left( {0.3} \right){{\left( {0.2} \right)}^2}{{\left( {0.8} \right)}^6}}}\end{aligned}\)

\(\xi \left( {0.1|x} \right) = 0.5418\)

and\(\xi \left( {0.2|x} \right)\)can be calculated as

\(\begin{aligned}\xi \left( {0.2|x} \right) = 1 - \xi \left( {0.1|x} \right)\\ = 1 - 0.5418\\ = 0.4582\end{aligned}\)

Hence Posterior pf of \(\theta \) is \(\xi \left( {0.1|x} \right) = 0.5418\)and \(\xi \left( {0.2|x} \right) = 0.4582\)