91影视

An outcome is a possible result of an experiment or trial in probability theory. Each conceivable experiment outcome is distinct, and alternative outcomes are mutually exclusive. A sample space is made up of all of the possible results of an experiment.

From the exercise\(26\), the following events can be taken, \({A_i}\)- "system has the defect of type \(i\)",\(i = 1,2,3\), and the following probabilities will be

\(\begin{array}{l}P\left( {{A_1}} \right) &=& 0.12,\;\;\;P\left( {{A_2}} \right) &=& 0.07,\;\;\;P\left( {{A_3}} \right) &=& 0.05\\P\left( {{A_1} \cup {A_2}} \right) &=& 0.13,\;\;\;P\left( {{A_1} \cup {A_3}} \right) &=& 0.14\\P\left( {{A_2} \cup {A_3}} \right) &=&0.1,\;\;\;P\left( {{A_1} \cap {A_2} \cap {A_3}} \right) &=& 0.01\end{array}\)

It is given that the system is having a type \(1\) defect (event\({A_1}\)), the probability that it has a type \(2\) defect (event\({A_2}\)) is calculated using:

The conditional probability of A given that the event B has occurred, for which \(P(B) > 0\), is

\(P(A\mid B) = \frac{{P(A \cap B)}}{{P(B)}}\)

for any two events A and B.

Therefore,

\(P\left( {{A_2}\mid {A_1}} \right) = \frac{{P\left( {{A_2} \cap {A_1}} \right)}}{{P\left( {{A_1}} \right.}}\mathop = \limits^{(1)} \frac{{0.06}}{{0.12}} = 0.5\)

(1): the \(P\left( {{A_2} \cap {A_1}} \right)\)is

\(\begin{array}{c}P\left( {{A_2} \cap {A_1}} \right) &=& P\left( {{A_1}} \right) + P\left( {{A_2}} \right) - P\left( {{A_1} \cup {A_2}} \right)\\ &=& 0.12 + 0.07 - 0.13\\ &=& 0.06\end{array}\)

For every two events A and B

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

It is given that the system is having a type \(1\) defect, the probability that it has all three types of defects is the conditional probability of the intersection of \({A_1},{A_2}\), and \({A_3}\)given that the event \({A_1}\)has occurred

\(\begin{array}{c}P\left( {{A_1} \cap {A_2} \cap {A_3}\mid {A_3}} \right) &=& \frac{{P\left( {\left( {{A_1} \cap {A_2} \cap {A_3}} \right) \cap {A_3}} \right)}}{{P\left( {{A_3}} \right)}}\\ &=& \frac{{P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)}}{{P\left( {{A_3}} \right)}}\\ &=& \frac{{0.01}}{{0.12}}\\ = 0.0833\end{array}\)

where the probabilities we needed are given in exercise \(26\).

It is given that the system is having at least one type of defect, denoted as A, the probability that it has exactly one type of defect, denoted as B, is

\(P(B\mid A) = \frac{{P(B \cap A)}}{{P(A)}}\)

We have the \(P\left( {{A_1} \cap {A_2}} \right)\)is

\(P\left( {{A_1} \cap {A_2}} \right) = P\left( {{A_1}} \right) + P\left( {{A_2}} \right) - P\left( {{A_1} \cup {A_2}} \right) = 0.12 + 0.07 - 0.13 = 0.06\)

also \(P\left( {{A_1} \cap {A_3}} \right)\)is

\(P\left( {{A_1} \cap {A_3}} \right) = P\left( {{A_1}} \right) + P\left( {{A_3}} \right) - P\left( {{A_1} \cup {A_3}} \right) = 0.12 + 0.05 - 0.14 = 0.03\)

and \(P\left( {{A_2} \cap {A_3}} \right)\)is

\(P\left( {{A_2} \cap {A_3}} \right) = P\left( {{A_2}} \right) + P\left( {{A_3}} \right) - P\left( {{A_2} \cup {A_3}} \right) = 0.07 + 0.05 - 0.1 = 0.02.\)

The probability that system has at least one type of defect is the union the three given events

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup {A_3}} \right)\mathop = \limits^{(1)} P\left( {{A_1}} \right) + P\left( {{A_2}} \right) + P\left( {{A_3}} \right) - P\left( {{A_1} \cap {A_2}} \right) - P\left( {{A_1} \cap {A_3}} \right) - P\left( {{A_2} \cap {A_3}} \right) + P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.12 + 0.07 + 0.05 - 0.06 - 0.03 - 0.02 + 0.01\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.14\end{array}\)

(1): here we used proposition:

For every three events A, B and C the following is true

Notice that intersection of events A and B is only B, because "exactly one" is contained in "at least one"\((B \subset A)\). Therefore, now we need only to find \(P(B)\). Event \({\rm{B}}\)is union of three disjoint event -\({A_1} \cap A_2^\prime \cap A_3^\prime ,A_1^\prime \cap {A_2} \cap A_3^\prime ,and\;A_1^\prime \cap A_2^\prime \cap {A_3}\), where they mean, respectively, only $A_{1}$ occurred and not \({A_2},{A_3}\), only \({A_2}\)occurred and not \({A_1},{A_3}\), only \({A_3}\)occurred and not \({A_1},{A_2}\)(which actually means exactly one type of defect occurred).

Therefore

\(\begin{array}{c}P(B) = P\left( {\left( {{A_1} \cap A_2^\prime \cap A_3^\prime } \right) \cup \left( {A_1^\prime \cap {A_2} \cap A_3^\prime } \right) \cup \left( {A_1^\prime \cap A_2^\prime \cap {A_3}} \right)} \right)\\\mathop = \limits^{(1)} P\left( {{A_1} \cap A_2^\prime \cap A_3^\prime } \right) + P\left( {A_1^\prime \cap {A_2} \cap A_3^\prime } \right) + P\left( {A_1^\prime \cap A_2^\prime \cap {A_3}} \right)\end{array}\)

(1): here we used the fact that they are disjoint.

We will calculate each intersection separately.

First:

\(\begin{array}{l}P\left( {{A_1} \cap A_2^\prime \cap A_3^\prime } \right)\mathop = \limits^{(1)} P\left( {{A_1}} \right) - P\left( {{A_1} \cap {A_2}} \right) - P\left( {{A_1} \cap {A_3}} \right) + P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.12 - 0.06 - 0.03 + 0.01\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.04\end{array}\)

Second:

\(\begin{array}{l}P\left( {A_1^\prime \cap {A_2} \cap A_3^\prime } \right)\mathop = \limits^{(1)} P\left( {{A_2}} \right) - P\left( {{A_1} \cap {A_2}} \right) - P\left( {{A_2} \cap {A_3}} \right) + P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.07 - 0.06 - 0.02 + 0.01\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0\end{array}\)

Third:

\(\begin{array}{l}P\left( {A_1^\prime \cap A_2^\prime \cap {A_3}} \right)\mathop = \limits^{(1)} P\left( {{A_3}} \right) - P\left( {{A_1} \cap {A_3}} \right) - P\left( {{A_2} \cap {A_3}} \right) + P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.05 - 0.03 - 0.02 + 0.01\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.01\end{array}\)

So, the probability of event B is

\(P(B) = 0.04 + 0 + 0.01 = 0.05\)

Finally, the probability of \(P(B\mid A)\)is

\(P(B\mid A) = \frac{{P(B \cap A)}}{{P(A)}} = \frac{{P(B)}}{{P(A)}} = \frac{{0.05}}{{0.14}} = 0.3571\)

Given that the system has both of the first two types of defects, the probability that it does not have the third type of defect is conditional probability of \(A_3^\prime \)given that the event \({A_1} \cap {A_2}\)has occurred

\(\begin{array}{c}P\left( {A_3^\prime \mid {A_1} \cap {A_2}} \right) = \frac{{P\left( {{A_1} \cap {A_2} \cap A_3^\prime } \right)}}{{P\left( {{A_1} \cap {A_2}} \right)}}\mathop = \limits^{(1)} \frac{{P\left( {{A_1} \cap {A_2}} \right) - P\left( {{A_1} \cap {A_2} \cap {A_3}} \right)}}{{P\left( {{A_1} \cap {A_2}} \right)}}\\\mathop = \limits^{(2)} \frac{{0.06 - 0.01}}{{0.06}}\\ = 0.833,\end{array}\)