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\({X_2} = 0,1,2,3\);

- when\({X_1} = 3\)or\({X_2} = 0,1,2\).

So, the probability of accepting is

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

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

\(\begin{array}{l}P({\rm{ accepted }}) = P\left( {{X_1} = 0{\rm{ or }}1} \right) + P\left( {{X_1} = 2} \right) \cdot P\left( {{X_2} = 0,1,2,3} \right)\\ + P\left( {{X_1} = 3} \right) \times P\left( {{X_2} = 0,1,2} \right) = \\\sum\limits_{y = 0}^1 b (y;50,p) + b(2;50,p) \times \sum\limits_{y = 0}^3 b (y;50,p) + b(3;50,p) \times \sum\limits_{y = 0}^2 b (y;n,p)\end{array}\)

For\(p = 0.01,0.05,0.1\), respectively, the probabilities of acceptance are

\(\begin{array}{l}\left. {P\left( {{A_1}} \right) = 0.9106 + 0.0756 \times 0.9984} \right) + 0.0122 \times 0.9862 = 0.9981\\P\left( {{A_2}} \right) = 0.2794 + 0.2611 \times 0.7604 + 0.2199 \times 0.5405 = 0.5968\\\left. {P\left( {{A_3}} \right) = 0.0338 + 0.0779 \times 0.2503} \right) + 0.1386 \times 0.1117 = 0.0688\end{array}\)