1 An industrial process produces components. Some of the components contain faults. The number of faults in a component is modelled by the random variable \(X\) with probability function $$\mathrm { P } ( X = x ) = \theta ( 1 - \theta ) ^ { x } \quad \text { for } x = 0,1,2 , \ldots$$ where \(\theta\) is a parameter with \(0 < \theta < 1\). The numbers of faults in different components are independent.
A random sample of \(n\) components is inspected. \(n _ { 0 }\) are found to have no faults, \(n _ { 1 }\) to have one fault and the remainder \(\left( n - n _ { 0 } - n _ { 1 } \right)\) to have two or more faults.

1. Find \(\mathrm { P } ( X \geqslant 2 )\) and hence show that the likelihood is $$\mathrm { L } ( \theta ) = \theta ^ { n _ { 0 } + n _ { 1 } } ( 1 - \theta ) ^ { 2 n - 2 n _ { 0 } - n _ { 1 } }$$
2. Find the maximum likelihood estimator \(\hat { \theta }\) of \(\theta\). You are not required to verify that any turning point you locate is a maximum.
3. Show that \(\mathrm { E } ( X ) = \frac { 1 - \theta } { \theta }\). Deduce that another plausible estimator of \(\theta\) is \(\tilde { \theta } = \frac { 1 } { 1 + \bar { X } }\) where \(\bar { X }\) is the sample mean. What additional information is needed in order to calculate the value of this estimator?
4. You are given that, in large samples, \(\tilde { \theta }\) may be taken as Normally distributed with mean \(\theta\) and variance \(\theta ^ { 2 } ( 1 - \theta ) / n\). Use this to obtain a \(95 \%\) confidence interval for \(\theta\) for the case when 100 components are inspected and it is found that 92 have no faults, 6 have one fault and the remaining 2 have exactly four faults each.