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.

Given:

\({\rm{P( disease ) = 0}}{\rm{.05}}\)

In 98 percent of cases, the disease is correctly recognised, and the test is thus positive:

\(\begin{array}{c}{\rm{P( positive}}\mid \rm disease ) &=& 98\% \\ \rm &=& 0{\rm{.98}}\end{array}\)

In 99 percent of cases, the absence of the disease is correctly recognised, and the test is consequently negative:

\(\begin{array}{c}{\rm{P( not positive}}\mid \rm no disease ) &=& 99\% \\ \rm &=& 0{\rm{.99}}\end{array}\)

Complement rule:

\({\rm{P(notA) = 1 - P(A)}}\)

General multiplication rule:

\(\begin{array}{c}\rm P(A and B) &=& P(A)*P(B\mid {\rm{A)}}\\ \rm &=& P(B)*P(A \mid {\rm{B)}}\end{array}\)

For discontinuous or mutually exclusive occurrences, use the following addition rule:

\({\rm{P(A or B) = P(A) + P(B)}}\)

Multiplication rule for independent events:

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

Definition Conditional probability:

\({\rm{P(B}}\mid {\rm{A) = }}\frac{{{\rm{P(A and B)}}}}{{{\rm{P(A)}}}}\)

Use the complement rule:

\(\begin{array}{c} \rm P( No disease ) &=& 1 - P( disease ) \\ \rm &=& 1 - 0 \rm .05 &=& 0{\rm{.95}}\\{\rm{P( detect }}\mid \rm no disease ) &=& 1 - P( not positive \mid \rm no disease )\\ \rm &=& 1 - 0{\rm{.99}}\\ \rm &=& 0{\rm{.01}}\end{array}\)

Use the general multiplication rule:

\(\begin{array}{c}\rm P( positive and disease ) &=& P( disease )*P( positive \mid {\rm{ disease )}}\\ \rm &=& 0{\rm{.05*0}}{\rm{.98}}\\ \rm &=& 0{\rm{.049}}\end{array}\)

\(\begin{array}{c}{\rm{P}}\left( {{\rm{ positive and no disease}}} \right)\rm &=& P\left( {{\rm{no disease}}} \right){\rm{*P(positive}}\mid {\rm{no disease )}}\\ \rm &=& 0{\rm{.950}}{\rm{.01}}\\ \rm &=& 0{\rm{.0095}}\end{array}\)

Add the corresponding probabilities:

\(\begin{array}{c} \rm P( positive ) &=& P( positive and disease ) + P( positive and no disease )\\ \rm &=& 0{\rm{.049 + 0}}{\rm{.0095}}\\ \rm &=& 0{\rm{.0585}}\end{array}\)

Because the two tests are independent, we can apply the multiplication rule to both. We're going to assume

\(\begin{array}{c}\rm P( positive twice ) &=& P( positive )*P( positive )\\ \rm &=& 0{\rm{.0585*0}}{\rm{.0585}}\\ \rm &=& 0{\rm{.00342225}}\end{array}\)

P(disease and positive twice )=P( positive and disease ) \(*\)P(positive and disease )

\(\begin{array}{c}\rm &=& 0{\rm{.049*0}}{\rm{.049}}\\\rm &=& 0{\rm{.002401}}\end{array}\)

Use the following conditional probability definition:

\(\begin{array}{c}{\rm{P( disease }}\mid \rm positive twice ) &=& \frac{{{\rm{P( disease and positive twice )}}}}{{{\rm{P( positive twice )}}}}\\ &=& \frac{{0.002401}}{{0.00342225}} \approx 0.7016\\ \rm &=& 70{\rm{.16\% }}\end{array}\)

Therefore , the selected individual has the disease

\(\begin{array}{c}{\rm{P( disease }}\mid {\rm{ positive twice )}} \rm &=& 0{\rm{.7016}}\\ \rm &=& 70{\rm{.16\% }}\end{array}\)