91影视

Independence If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

Denote events

\({{\rm{A}}_{\rm{1}}}\){flaw detected during first fixation}

\({{\rm{A}}_{\rm{2}}}\){flaw detected during second fixation}

As given in the exercise, denote

\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = p}}\)

(a):

By the end of the second fixation, denote event D as "detected." Once a flaw is found, the fixation sequence ends, therefore event \({\rm{D}}\)can be represented as the union of two events: \({{\rm{A}}_{\rm{1}}}\) (if the flaw is detected in the first fixation), and \(A_1^\prime \cap {{\rm{A}}_{\rm{2}}}\) (if the defect is detected in the second fixation) (it is not detected in first fixation but in second). Keep in mind that we thought an item had a fault (so the statement above is true). The following is an example of what you can do with this.

\(\begin{align}\text P(D)=&P \left( {{A}_{1}}\cup \left( A_{1}^{\prime }\cap {{A}_{2}} \right) \right) \\ \text =&P \left( {{\text{A}}_{\text{1}}} \right)\text{+P}\left( A_{1}^{\prime }\cap {{A}_{2}} \right)............(1) \\ \text =&P \left( {{\text{A}}_{\text{1}}} \right)\text{+P}\left( A_{1}^{\prime } \right)\cdot P\left( {{A}_{2}} \right).........(2) \\ \text =& p+(1-p)*p............\text{(3)} \\ \text =& p*(2-p) \\ \end{align}\)

(1): the events are disjoint,

(2): events \(A_1^\prime \) and \({{\rm{A}}_{\rm{2}}}\)are independent because events \({{\rm{A}}_{\rm{1}}}\)and \({{\rm{A}}_{\rm{2}}}\)are independent (see proposition below). Since they are

independent, we can use the multiplication rule given below,

(3): for any event A, \({\rm{P(A) + P}}\left( {{A^\prime }} \right){\rm{ = 1}}\), therefore, \({\rm{P}}\left( {A_1^\prime } \right){\rm{ = 1 - p}}\).

Multiplication Property: Two events \({\rm{A}}\)and \({\rm{B}}\)are independent if and only if

\(P(A \cap B){\rm{ = P(A)*P(B)}}\)

Proposition: If \({\rm{A}}\)and \({\rm{B}}\)are independent, then

- \({A^\prime }\)and \({\rm{B}}\)are independent;

- \({A^\prime }\)and \({B^\prime }\)are independent;

- \({\rm{A}}\)and \({B^\prime }\)are independent.

Thus, The probability that it is detected by the end of the second fixation is \({\rm{P(D) = p*(2 - p)}}\);

(b):

Denote events

\({{\rm{A}}_{\rm{1}}}\){flaw detected during 1^st fixation}

\({{\rm{A}}_{\rm{2}}}\){flaw detected during 2^nd fixation }

\({A_n} = \left\{ {} \right.\)flaw detected during 1^st fixation} .

As given in the exercise, denote

\({\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{ = p}}\)

for \({\rm{i = 1,2, \ldots ,n}}\).

Remember that we know an object has a problem and that the series of fixations ends when a flaw is found, just as we did in (a). We can represent event E= By the end of the nth fixation , will be identified as a union of \({\rm{n}}\) disjoint events.

\(\begin{align}P(E)=& P\left( {{A}_{1}}\cup \left( A_{1}^{\prime }\cap {{A}_{2}} \right)\cup \left( A_{1}^{\prime }\cap A_{2}^{\prime }\cap A_{3}^{\prime } \right)\cup \ldots \cup \left( A_{1}^{\prime }\cap A_{2}^{\prime }\cap \ldots \cap A_{n-1}^{\prime }\cap {{A}_{n}} \right) \right)..............(1) \\ =& P\left( {{A}_{1}} \right)+P\left( A_{1}^{\prime }\cap {{A}_{2}} \right)+\ldots +P\left( A_{1}^{\prime }\cap A_{2}^{\prime }\cap \ldots \cap A_{n-1}^{\prime }\cap {{A}_{n}} \right)..............(2) \\ =& P\left( {{A}_{1}} \right)+P\left( A_{1}^{\prime } \right)\cdot P\left( {{A}_{2}} \right)+P\left( A_{1}^{\prime } \right)\cdot P\left( A_{2}^{\prime } \right)\cdot P\left( {{A}_{3}} \right)+\ldots +P\left( A_{1}^{\prime } \right)\cdot P\left( A_{2}^{\prime } \right)\cdot \ldots \cdot P\left( A_{n-1}^{\prime } \right)\cdot P\left( {{A}_{n}} \right)..............(3) \\ =& \text{p+(1-p)*p+(1-p}{{\text{)}}^{\text{2}}}\text{*p++(1-p}{{\text{)}}^{\text{n-1}}}\text{*p} \\ \text =& p*\left( \text{1+(1-p)+(1-p}{{\text{)}}^{\text{2}}}\text{++(1-p}{{\text{)}}^{\text{n-1}}} \right) \\ \text =&p* \frac{\text{1-(1-p}{{\text{)}}^{\text{n}}}}{\text{1-(1-p)}}..................(4) \\ \end{align}\)

\(=1-{{(1-p)}^{n}}\)

(1): note the explanations in (a) and (b) at the start of (b),

(2): the sequence of events is disjointed,

(3): the following multiplication rule and statement,

(4): the following geometric series.

Proposition: If \({\rm{A}}\)and \({\rm{B}}\)are independent, then

- \({A^\prime }\)and \({\rm{B}}\)are independent;

- \({A^\prime }\)and \({B^\prime }\)are independent;

- \({\rm{A}}\)and \({B^\prime }\)are independent.

This is also true for \({\rm{n}}\) events (any combination of events \({{\rm{A}}_{\rm{i}}}\)and their complements is treated as a separate event).

Multiplication Property:

For events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}{\rm{,n}} \in {\rm{N}}\)we say that they are mutually independent if

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

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

Formula can be used to determine the sum of finite geometric series.

where \({\rm{n}}\)is number of terms, \({a_1}\) is the first term and \({\rm{q}}\) is the common ratio. In our case \({{\rm{a}}_{\rm{1}}}{\rm{ = 1}}\) and \({\rm{q = 1 - p < 1}}\).

Thus, The probability that a flawed item will pass inspection is \({\rm{P(E) = 1 - (1 - p}}{{\rm{)}}^{\rm{n}}}{\rm{;}}\)

(c):

Denote event \({{\rm{N}}_{\rm{3}}}{\rm{ = \{ }}\)the flas has no been detected in 3 fixations \(\} \). This event is intersection of events \(A_1^\prime \) (no detected in first), \(A_2^\prime \) (not detected in second), and \(A_3^\prime \) (not detected in third).

Therefore

\(\begin{aligned}{\rm{P(}}{{\rm{N}}_{\rm{3}}} \rm) =& P({{A'}_1} \cap {{A'}_2} \cap {{A'}_3}) \rm &=& P({{A'}_1})*{\rm{P}}({{A'}_2})*{\rm{P}}({{A'}_3})\\ \rm =& (1 - p{{\rm{)}}^{\rm{3}}}\end{aligned}\)

(1): we employ the multiplication rule, and events are independent of one another \({A'_1},{A'_2}\) and \({A'_3}\)and the propositions mentioned in (b).

Thus, The probability that a flawed item will pass inspection is \({\rm{P(}}{{\rm{N}}_{\rm{3}}}{\rm{) = (1 - p}}{{\rm{)}}^{\rm{3}}}\);

(d): Denote event A = { randomly chosen item passes inspection } It is possible for an item to be defective and pass or not be flawed and pass, which is a combination of occurrences NF = { not flawed and passes } and FP={ flawed and passes },

Therefore

\(\begin{align}\text P(A)=&P(NF \cup \text{FP})=\text{P(NF)+P(FP)}........\text{(1)} \\ \text =&P(NF)+P(F)*P(P\!\!|\!\!\text{ F)}...........\text{(2)} \\ \text =&0 \text{.9+0}\text{.1*(1-p}{{\text{)}}^{\text{3}}}.................(3) \\ \end{align}\)

(1): events NF and FP are disjoint;

(2): here we use the multiplication rule given below;

(3): Because the likelihood of a randomly chosen object being flawed is 0.1, the probability of it not being flawed is \(1 - 0.1 = 0.9\). We also utilise what we computed in (c) - P(P|F) (recall that in (c), we assumed we had a flawed item, thus we calculated the conditional probability of an event-flawed product passing provided that the event-flawed product has occurred).

The Multiplication Rule

\({\rm{P(A}} \cap {\rm{B) = P(A|B)*P(B) }}\)

Thus, The probability that a randomly chosen item will pass inspection is \({\rm{P(A) = 0}}{\rm{.9 + 0}}{\rm{.1*(1 - p}}{{\rm{)}}^{\rm{3}}}{\rm{ }}\);

(e): Given that the event P (passes) has occurred, we must calculate the conditional probability of F (flawed). This can be calculated using the definition of conditional probability.

\({\rm{P}}\left( {\rm{B}} \right){\rm{ > 0}}\)is the conditional probability of A given the occurrence of event B.

\({\rm{P(A|B) = }}\frac{{{\rm{P(A}} \cap {\rm{B)}}}}{{{\rm{P}}({\rm{B}})}}{\rm{ }}\)

for any two event A and B.

The following is true

\(\begin{align}\text{P(F }\!\!|\!\!\text P)=& \frac{\text{P(F}\cap \text{P)}}{\text{P(P})} \\ =&\frac{\text{0}\text{.1*(1-p}{{\text{)}}^{\text{3}}}}{\text{0}\text{.9+0}\text{.1*(1-p}{{\text{)}}^{\text{3}}}}\text{,}.........\text{(1)} \end{align}\)

(1): we have calculated the probabilities in (d).

For\({\rm{p = 0}}{\rm{.5}}\), we have

\(\begin{aligned} \rm P(F|P) =&\frac{{{\rm{0}}{\rm{.1*(1 - p}}{{\rm{)}}^{\rm{3}}}}}{{{\rm{0}}{\rm{.9 + 0}}{\rm{.1*(1 - p}}{{\rm{)}}^{\rm{3}}}}}\\ \rm =& \frac{{{\rm{0}}{\rm{.1*(1 - 0}}{\rm{.5}}{{\rm{)}}^{\rm{3}}}}}{{{\rm{0}}{\rm{.9 + 0}}{\rm{.1*(1 - 0}}{\rm{.5}}{{\rm{)}}^{\rm{3}}}}}\\ \rm =& 0{\rm{.0137}}{\rm{.}}\end{aligned}\)

Thus, The probability that it is actually flawed is \({\rm{P(F|P) = 0}}{\rm{.0137}}{\rm{.}}\)