÷ÈÓ°Ö±²¥

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

\(\begin{array}{l}{\rm{A = \{ older pump fails \} }}\\{\rm{B = \{ newer pump fails \} }}{\rm{.}}\end{array}\)

We are given

\(\begin{array}{l}{\rm{P}}\left( {A \cap {B^\prime }} \right){\rm{ = 0}}{\rm{.1}}\\{\rm{P}}\left( {B \cap {A^\prime }} \right){\rm{ = 0}}{\rm{.05}}.\end{array}\)

Or equally

\(\begin{array}{l}{\rm{P(A) - P}}(A \cap B){\rm{ = 0}}{\rm{.1}}\\{\rm{P(B) - P}}(A \cap B){\rm{ = 0}}{\rm{.05}}\end{array}\)

We must locate \({\rm{P(A}} \cap {\rm{B)}}\), which we may do by employing the multiplication property.

Two occurrences \({\rm{A}}\)and \({\rm{B}}\)are independent if and only if they have the same multiplication property.

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

Denote with \({\rm{x = P(A}} \cap {\rm{B)}}\), we have

\(\begin{array}{c}{\rm{P(A) = 0}}{\rm{.1 + xP(B)}}\\{\rm{ = 0}}{\rm{.05 + x}}\end{array}\)

The following is correct when using the multiplication property.

\(P(A \cap B){\rm{ = P(A)*P(B) = (0}}{\rm{.1 + x)*(0}}{\rm{.05 + x)}}\)

from which we get

\({\rm{x = 0}}{\rm{.005 + 0}}{\rm{.1*x + 0}}{\rm{.05*x + }}{{\rm{x}}^{\rm{2}}}{\rm{.}}\)

This can be expressed as a quadratic formula.

\({{\rm{x}}^{\rm{2}}}{\rm{ - 0}}{\rm{.85*x + 0}}{\rm{.005 = 0,}}\)

where \({\rm{a = 1,b = - 0}}{\rm{.85}}\), and \({\rm{c = 0}}{\rm{.005}}\). This implies that solutions to the quadratic equation are

\(\begin{array}{l}{{\rm{x}}_{{\rm{12}}}}{\rm{ = }}\frac{{{\rm{ - b \pm }}\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} }}{{{\rm{2a}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.85 \pm }}\sqrt {{\rm{0}}{\rm{.8}}{{\rm{5}}^{\rm{2}}}{\rm{ - 4*1*0}}{\rm{.005}}} }}{{{\rm{2*1}}}}\end{array}\)

where first solution is

\(\begin{array}{l}{{\rm{x}}_{\rm{1}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.85 + }}\sqrt {{\rm{0}}{\rm{.8}}{{\rm{5}}^{\rm{2}}}{\rm{ - 4*0}}{\rm{.0}}} }}{{\rm{2}}}\\{x_{\rm{1}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.85 + }}\sqrt {{\rm{0}}{\rm{.7025}}} }}{{\rm{2}}} \Leftrightarrow {{\rm{x}}_{\rm{1}}}{\rm{ = 0}}{\rm{.8441}}{\rm{.}}\end{array}\)

and the second solutions

\({{\rm{x}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.85 - }}\sqrt {{\rm{0}}{\rm{.8}}{{\rm{5}}^{\rm{2}}}{\rm{ - 4*0}}{\rm{.005}}} }}{{\rm{2}}} \Leftrightarrow {{\rm{x}}_{\rm{2}}}{\rm{ = 0}}{\rm{.0059}}{\rm{.}}\)

Assume that the true chance is \({{\rm{x}}_{\rm{1}}}\). From

Proposition: For every two events \({\rm{A}}\)and \({\rm{B}}\)

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

we have

\(\begin{array}{c}P(A \cup B) = 0.1 + 0.8441 + 0.05 + 0.8441 - 0.8441\\ = 0.9941\end{array}\)

We can get \({{\rm{x}}_{\rm{2}}}\) if we believe true probability is

\(\begin{array}{c}P(A \cup B) = 0.1 + 0.0059 + 0.05 + 0.0059 - 0.0059\\ = 0.1559.\end{array}\)

Both are viable options for completing the exercise.

NOTE: For the scenario when \({{\rm{x}}_{\rm{1}}}\)is the true probability of intersection, we have that \(P(A \cup B) > 1\), which is not feasible, hence the only answer is \({{\rm{x}}_{\rm{2}}}\). Otherwise, it's possible that the creator intended for the exercise to have two options.

Therefore, the probability is \(\begin{array}{l}{\rm{P(}}A \cap B){\rm{ = 0}}{\rm{.0059; or }}\\{\rm{P(}}A \cap B){\rm{ = 0}}{\rm{.8441}}{\rm{.}}\end{array}\)