Question

One of the assumptions underlying the theory of control charting is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction.

Short answer

The solution is

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{10}}} \right) = 0.401\\P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{25}}} \right) = 0.723\end{array}\)

Step-by-step solution

Introduction

The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

Explanation

Denote events:

\({{\rm{A}}_{\rm{i}}}{\rm{ = \{ }}\)point i error was signaled incorrectly\({\rm{\} ,i = 1,2, \ldots ,25}}\).

The probabilities of each of the events is the same and it is \({\rm{0}}{\rm{.05}}\) (given in the exercise).

We are asked to find the probability that at least one of \({\rm{10}}\) successive points indicate a problem when in fact the process is operating correctly which is the union of events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{{\rm{10}}}}\) (at least one)

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{10}}} \right)\\\mathop = \limits^{(1)} P\left( {{{\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{10}^\prime } \right)}^\prime }} \right)\\\mathop = \limits^{(2)} 1 - P\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{10}^\prime } \right)\\\mathop = \limits^{(3)} 1 - P\left( {A_1^\prime } \right) \cdot P\left( {A_2^\prime } \right) \cdot \ldots \cdot P\left( {A_{10}^\prime } \right)\\\mathop = \limits^{(4)} 1 - (1 - 0.05) \cdot (1 - 0.05) \cdot \ldots \cdot (1 - 0.05)\\ = 1 - {0.95^{10}}\\ = 0.401\end{array}\)

(1): here we use De Morgan's Law,

(2): for any event\(A,P\left( {{A^\prime }} \right) + P(A) = 1\),

(3): the events (points) are independent, so we can use the multiplication property given below,

Explanation of multiplication property

(4): using\(A,P\left( {{A^\prime }} \right) + P(A) = 1\).

Multiplication Property:

For events \({A_1},{A_2}, \ldots ,{A_n},n \in \mathbb{N}\)we say that they are mutually independent if

\(\begin{array}{l}P\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right)\\ = P\left( {{A_{{i_1}}}} \right) \cdot P\left( {{A_{{i_2}}}} \right) \cdot \ldots \cdot P\left( {{A_{{i_k}}}} \right)\end{array}\)

for every\(k \in \{ 2,3, \ldots ,n\} \), and every subset of indices\({{\rm{i}}_{\rm{1}}}{\rm{,}}{{\rm{i}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{i}}_{\rm{k}}}\).

Given \({\rm{25}}\) successive points, similarly we obtain

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{25}}} \right)\\\mathop = \limits^{(1)} P\left( {{{\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{25}^\prime } \right)}^\prime }} \right)\\\mathop = \limits^{(2)} 1 - P\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{25}^\prime } \right)\\\mathop = \limits^{(3)} 1 - P\left( {A_1^\prime } \right) \cdot P\left( {A_2^\prime } \right) \cdot \ldots \cdot P\left( {A_{25}^\prime } \right)\\\mathop = \limits^{(4)} 1 - (1 - 0.05) \cdot (1 - 0.05) \cdot \ldots \cdot (1 - 0.05)\\ = 1 - {0.95^{25}}\\ = 0.723.\end{array}\)

(1): here we use De Morgan's Law,

(2): for any event\(A,P\left( {{A^\prime }} \right) + P(A) = 1\),

(3): the events (points) are independent, so we can use the multiplication property given above,

(4): using\(A,P\left( {{A^\prime }} \right) + P(A) = 1\)

Therefore, the result is

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{10}}} \right) = 0.401\\P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{25}}} \right) = 0.723\end{array}\)