10% of the population in a town is HIV+ . A new diagnostic kit for HIV detection is available; this kit correctly identifies HIV+ individuals 95% of the time, and HIV− individuals 89% of the time. A particular patient is tested using this kit and is found to be positive. The probability that the individual is actually positive is _______
Concept:
Bayes' Theorem:
Let E1, E2, ….., En be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E1 or E2 or … or En such that P(A) ≠ 0. Then
\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\;{E_i}} \right)}}{{\mathop \sum \nolimits_{i\; = 1}^n P\left( {{E_i}} \right) \times P\left( {A\;|\;{E_i}} \right)}},\;i = 1,\;2,\; \ldots .\;,\;n\)
Calculation:
The patient is actually HIV +ve.
So probability of that = P(he belongs to HIV +ve population) * P( HIV +ve test gives correct result given he is HIV +ve).
∴ Probabilty P(A) = 0.1 and P(E/A) = 0.95
The next case is a patient is HIV -ve but the test is showing HIV +ve i.e. test shows wrong result.
So, P(B) = 0.9 and P(E/B) = 1 - 0.89 = 0.11
So By Bayes' Theorem,
P(A/E) i.e. P(actually is HIV +ve) = \(\frac{{\left\{ {{\rm{P}}\left( {\rm{A}} \right){\rm{\times\;P}}\left( {\frac{{\rm{E}}}{{\rm{A}}}} \right)} \right\}}}{{\left[ {{\rm{\;}}\left( {{\rm{P}}\left( {\rm{A}} \right){\rm{\;\times\;P}}\left( {{\rm{E}}/{\rm{A}}} \right)} \right){\rm{\;}} + {\rm{\;}}\left( {{\rm{P}}\left( {\rm{B}} \right){\rm{\;\times\;P}}\left( {{\rm{E}}/{\rm{B}}} \right)} \right){\rm{\;}}} \right]}}{\rm{\;}}\)
\( = \frac{{0.1\;\times\;0.95}}{{\;\left[ {\left( {0.1\;\times\;0.95} \right)\; + \;\left( {0.9\;\times\;0.11} \right)} \right]}}\)
\( = \frac{{0.095}}{{\;0.095\; + \;0.099}}\)
= 0.095 / 0.194
= 95 / 194
= 0.4897
= 48.97 %
Important Questions from Conditional Probability
Two events A and B are such that P(not B) = 0.8, P(A ∪ B) = 0.5 and P(A|B) = 0.4. Then P(A) is equal to
For two events, A and B, it is given that \({\rm{P}}\left( {\rm{A}} \right) = \frac{3}{5},{\rm{\;P}}\left( {\rm{B}} \right) = \frac{3}{{10}}\) and \({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{2}{3}\) . If A̅ and B̅ are the complementary events of A and B, then what is P(A̅ | B̅) equal to?
For two mutually exclusive events A and B, P(A) = 0.2 and P (A̅ ∩ B) = 0.3. What is P (A|(A ∪ B)) equal to?
If an event B has occurred and has P(B) = 1, the conditional probability P(A|B) is equal to:
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
