91Ó°ÊÓ

Let X denote the number of defects on the selected roll of tape.

For any given value of λ, the p.f. of X is:

\(f\left( {X|\lambda } \right) = \frac{{\exp \left( { - \lambda } \right){\lambda ^x}}}{{x!}}\)for\(0,1,2...\)

Definition of the posterior pf is:

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

Where\({g_n}\left( x \right)\)is the marginal pdf of\({X_{1,}}...{X_n}\)

Hence As the role of tape selected at random found three defects and the mean\(\lambda \)is either1.0 or 1.5,

\(\begin{aligned}\xi \left( {1.0|X = 3} \right) = {\rm P}\left( {\lambda = 1.0|X = 3} \right)\\ = \frac{{\xi \left( {1.0} \right)f\left( {3|1.0} \right)}}{{\xi \left( {1.0} \right)f\left( {3|1.0} \right) + \xi \left( {1.5} \right)f\left( {3|1.5} \right)}}\end{aligned}\)

From the table of the Poisson distribution it is found that

\(f\left( {3|1.0} \right) = 0.0613\)and

\(f\left( {3|1.5} \right) = 0.1255\).

Therefore,

\(\begin{aligned}\xi \left( {1.0|X = 3} \right) = \frac{{\xi \left( {1.0} \right) \times 0.0613}}{{\xi \left( {1.0} \right)0.0613 + \xi \left( {1.5} \right) \times 0.1255}}\\ = 0.2456\end{aligned}\)

and \(\xi \left( {1.5|X = 3} \right)\)can be calculated as

\(\begin{aligned}\xi \left( {1.5|X = 3} \right) = 1 - \xi \left( {1.0|X = 3} \right)\\ = 1 - 0.2456\\ = 0.75544\end{aligned}\).

Hence Posterior pf of \(\lambda \) \(\xi \left( {1.0|X = 3} \right) = 0.2456\) and \(\xi \left( {1.5|X = 3} \right) = 0.75544\)