91Ó°ÊÓ

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.

(a):

Using the fact that events are mutually independent, we can use the multiplication property.

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

\({\rm{P}}\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{1}}}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{2}}}}}} \right){\rm{* \ldots *P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{k}}}}}} \right)\)

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

It's worth noting that the probability we're asked to calculate is the result of the intersection of all three events (all three components function properly)

\(\begin{align} P\left( {{A}_{1}}\cap {{A}_{2}}\cap {{A}_{3}} \right)\text =&0 \text{.95*0}\text{.98*0}\text{.8}..........\text{(1)} \\ \text =&0 \text{.7448} \end{align}\)

(1): multiplication property.

Thus, the multiplication of property \(P\left( {{A_1} \cap {A_2} \cap {A_3}} \right){\rm{ = 0}}{\rm{.7448}}.\)

We determined the chance of all three components working properly throughout the warranty period in (a), which is the complement of at least one component requiring service during the warranty period.

Therefore,

\(\begin{align}P\left[ {{\left( {{A}_{1}}\cap {{A}_{2}}\cap {{A}_{3}} \right)}^{\prime }} \right]\text =& 1-P\left( {{A}_{1}}\cap {{A}_{2}}\cap {{A}_{3}} \right)..........(1) \\ \text =& 1-0 \text{.7448} \\ \text =& 0 \text{.2552,} \end{align}\)

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

\({\rm{P}}\left( {{{\left( {{A_1} \cap {A_2} \cap {A_3}} \right)}^\prime }} \right){\rm{ = 0}}{\rm{.2552}}{\rm{.}}\)

The probability that all three components need service during the warranty period is probability of intersection of events \(A_1^\prime ,A_2^\prime \), and \(A_3^\prime \)

\(\begin{align}P\left( A_{1}^{\prime }\cap A_{2}^{\prime }\cap A_{3}^{\prime } \right)=& P\left( A_{1}^{\prime } \right)\cdot P\left( A_{2}^{\prime } \right)\cdot P\left( A_{3}^{\prime } \right)...........(1) \\ \text =& \left( \text{1-P}\left( {{\text{A}}_{\text{1}}} \right) \right)*\left( \text{1-P}\left( {{\text{A}}_{\text{2}}} \right) \right)*\left( \text{1-P}\left( {{\text{A}}_{\text{3}}} \right) \right) \\ \text =&0 \text{.05*0}\text{.02*0}\text{.2} \\ \text =&0 \text{.0002}\end{align}\)

(1) : We employ the multiplication property listed below (we can do this because events are assumed to be mutually independent).

Property of Multiplication:

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

\({\rm{P}}\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right) = {\rm{P}}\left( {{A_{{i_1}}}} \right){\rm{*}}{{\rm{A}}_{{{\rm{i}}_{\rm{2}}}}}{\rm{* \ldots *}}{{\rm{A}}_{{{\rm{i}}_{\rm{k}}}}}\)

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

We also use proposition: If \({\rm{A}}\)and \({\rm{B}}\)are independent, then

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

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

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

which is true for \({\rm{n}}\) events too (each combination of events \({{\rm{A}}_{\rm{i}}}\)and their complements are independent events). This is how we know that events \(A_1^\prime ,A_2^\prime \), and \(A_3^\prime \)are independent.

\(P\left( {A_1^\prime \cap A_2^\prime \cap A_3^\prime } \right) = 0.0002.\)

(d):

The probability that only the receiver needs service during the warranty period is probability of intersection of events \(A_1^\prime \) (receiver does not function properly), \({{\rm{A}}_{\rm{2}}}\) (speakers function properly) and \({{\rm{A}}_{\rm{3}}}\) (CD player function properly)

\(\begin{align}P\left( A_{1}^{\prime }\cap {{A}_{2}}\cap {{A}_{3}} \right)=& P\left( A_{1}^{\prime } \right)\text{*P}\left( {{\text{A}}_{\text{2}}} \right)\text{*P}\left( {{\text{A}}_{\text{3}}} \right)..........(1) \\ \text =& \left( \text{1-P}\left( {{\text{A}}_{\text{1}}} \right) \right)\text{*P}\left( {{\text{A}}_{\text{2}}} \right)\text{*P}\left( {{\text{A}}_{\text{3}}} \right) \\ \text =&0 \text{.05*0}\text{.98*0}\text{.8} \\ \text =&0 \text{.0392} \\ \end{align}\)

(1) : The multiplication property is used in this example (we can do that because we are given that events are mutually independent).

Property of Multiplication:

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

\({\rm{P}}\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right) = {\rm{P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{1}}}}}} \right){\rm{*P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{2}}}}}} \right){\rm{* \ldots *P}}\left( {{{\rm{A}}_{{{\rm{i}}_{\rm{k}}}}}} \right)\)

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

We also use proposition: If \({\rm{A}}\)and \({\rm{B}}\)are independent, then

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

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

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

which is true for \({\rm{n}}\)events too (each combination of events \({{\rm{A}}_{\rm{i}}}\)and their complements are independent events). This is how we know that events \(A_1^\prime ,{{\rm{A}}_{\rm{2}}}\), and \({{\rm{A}}_{\rm{3}}}\)are independent.

Thus , one reciver needs service\(P\left( {A_1^\prime \cap {A_2} \cap {A_3}} \right){\rm{ = 0}}{\rm{.0392}}{\rm{.}}\)

(e):

The probability that exactly one of the three components needs service during the warranty period is union of three disjoint events: \(P\left( {A_1^\prime \cap {A_2} \cap {A_3}} \right)\) (only receiver needs service), \(P\left( {{A_1} \cap A_2^\prime \cap {A_3}} \right)\) (only speakers need service), and \(P\left( {{A_1} \cap {A_2} \cap } \right.\left. {A_3^\prime } \right)\) (only CD player needs service).

Therefore,

\(\begin{align}P(\{\text{ exactly one needs service }\})=& P\left( A_{1}^{\prime }\cap {{A}_{2}}\cap {{A}_{3}} \right)+P\left( A_{1}^{\prime }\cap {{A}_{2}}\cap {{A}_{3}} \right)+P\left( A_{1}^{\prime }\cap {{A}_{2}}\cap {{A}_{3}} \right).........(1) \\ =&\left( 1-P\left( {{A}_{1}} \right) \right)\cdot P\left( {{A}_{2}} \right)\cdot P\left( {{A}_{3}} \right)+\left( 1-P\left( {{A}_{2}} \right) \right)\cdot P\left( {{A}_{1}} \right)\cdot P\left( {{A}_{3}} \right)+\left( 1-P\left( {{A}_{3}} \right) \right)\cdot P\left( {{A}_{1}} \right)\cdot P\left( {{A}_{2}} \right).........(2) \\ =&0.05\cdot 0.98\cdot 0.8+0.02\cdot 0.95\cdot 0.8+0.2\cdot 0.95\cdot 0.98 \\ =&0.0392+0.0152+0.1862 \\ =&0.2406, \end{align}\)

(1) : we use the fact that events are disjoint,

(2) : from the multiplication rule and proposition given below.

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

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

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

We also use 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.

which is true for \({\rm{n}}\)events too (each combination of events \({{\rm{A}}_{\rm{i}}}\)and their complements are independent events). This is how we know that events \(A_1^\prime {\rm{,}}{{\rm{A}}_{\rm{2}}}\), and \({{\rm{A}}_{\rm{3}}}\)are independent, events \({{\rm{A}}_{\rm{1}}}{\rm{,}}A_2^\prime \), and \({A_3}\)are independent and events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{,}}\)and \(A_3^\prime \)are independent.

Thus , the one needs service \({\rm{P(\{ exactly one needs service \} ) = 0}}{\rm{.2406}}{\rm{.}}\)

(f).

Given:

\(\begin{array}{l}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.95}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.98}}\\{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.80}}\end{array}\)

\({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{,}}{{\rm{A}}_{\rm{3}}}\)define the events that occur throughout the warranty period for an audio component.

However, because we don't know how long the warranty lasts or the likelihood that a component would fail within a month after it expires, we don't have enough data to calculate the indicated probability.

Thus, Cannot be determined