91Ó°ÊÓ

Consider the double-sampling plan. When will the lot be accepted? In the following cases:

- when\({X_1} = 0\)or\({X_1} = 1\);

- when\({X_1} = 2\)or\({X_2} = 0,1\);

- when\({X_1} = 3\)or\({X_2} = 0\),

This stands because

\({c_2} = {r_1} - 1 = 3\)

So, the probability of accepting is

\(\begin{array}{l}P( accepted ) = P\left( {{X_1} = 0 or 1} \right) + P\left( {{X_1} = 2} \right) \times P\left( {{X_2} = 0,1} \right)\\ + P\left( {{X_1} = 3} \right) \times P\left( {{X_2} = 0} \right)\end{array}\)

Use the following to compute the probabilities:

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{,{\rm{ otherwise }}}\end{array}} \right.\)

Cumulative Density Function cdf of binomial random variable\(X\)with parameters\(n\)and\(p\)is

\(B(x;n,p) = P(X \le x) = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\)

The probability is

\(\begin{array}{l}P( accepted ) = P\left( {{X_1} = 0 or 1} \right) + P\left( {{X_1} = 2} \right) \times P\left( {{X_2} = 0,1} \right)\\ + P\left( {{X_1} = 3} \right) \times P\left( {{X_2} = 0} \right)\end{array}\)

\( = \sum\limits_{y = 0}^1 b (y;50,p) + b(2;50,p) \times \sum\limits_{y = 0}^1 b (y;100,p) + b(3;50,p) \times b(0;100,p)\)

be careful when \(n = 100\)and when \(n = 50\left( {{n_1}} \right.\) and \(\left. {{n_2}} \right)\).

For\(p = 0.02,0.05,0.1\), respectively, the probabilities of acceptance are

\(\begin{array}{l}P\left( {{A_1}} \right) = 0.7358 + 0.1858 \cdot 0.4033 + 0.0607 \cdot 0.1326 = 0.8188\\P\left( {{A_2}} \right) = 0.2794 + 0.2611 \cdot 0.0371 + 0.2199 \cdot 0.0059 = 0.2904\\P\left( {{A_3}} \right) = 0.0338 + 0.0779 \cdot 0.0003 + 0.1386 \cdot 0 = 0.0038.\end{array}\)