# Baye's Theorem question

• Apr 12th 2008, 12:06 PM
lllll
Baye's Theorem question
the question goes as follows:

Medical case histories indicate that different illnesses may produce identical symptoms. Suppose that a particular set of symptoms, denoted $\displaystyle H$, occurs only when one of the tress illnesses, $\displaystyle I_1,\ 1_2, \ I_3$ occurs. Assume that the simultaneous occurrence of more ten one of these illnesses is impossible and that

$\displaystyle P(I_1) = 0.01, \ P(I_2) = 0.005, \ P(I_3) = 0.02$

$\displaystyle P(H|I_1) = 0.9, \ P(H|I_2) = 0.95, \ P(H|I_3) = 0.75$

assuming that an ill person exhibits the symptoms, $\displaystyle H$, what is the probability that the person has illness $\displaystyle I_1$?

solution:

$\displaystyle P(I_1|H) = \ ?$

$\displaystyle P(I_1|H) = \frac{H \cap I_1}{P(H)} \ = \ \frac{(H|I_1)(I_1)}{P(H)}$

To find $\displaystyle P(H) = \left( P(H \cap I_1) \ \bigcup \ P(H \cap \overline{I_1}) \right) \ \bigcup \ \left( P(H \cap I_2) \bigcup \ P(H \cap \overline{I_2}) \right) \ \bigcup$ $\displaystyle \left(P(H \cap I_3) \ \bigcup \ P(H \cap \overline{I_3}) \right)$

so $\displaystyle P(H) = (0.9)(0.01) +0 +(0.95)(0.05) +0+(0.75)(0.02) +0 = 0.1525$

therefore: $\displaystyle P(I_1|H) = \frac{(0.9)(0.1)}{0.1525} = 0.590164$

is this correct, or am I missing something?
• Apr 12th 2008, 12:19 PM
Plato
Quote:

Originally Posted by lllll
Medical case histories indicate that different illnesses may produce identical symptoms. Suppose that a particular set of symptoms, denoted $\displaystyle H$, occurs only when one of the tress illnesses, $\displaystyle I_1,\ 1_2, \ I_3$ occurs. Assume that the simultaneous occurrence of more than one of these illnesses is impossible and that
$\displaystyle P(I_1) = 0.01, \ P(I_2) = 0.005, \ P(I_3) = 0.02$
$\displaystyle P(H|I_1) = 0.9, \ P(H|I_2) = 0.95, \ P(H|I_3) = 0.75$
“Assume that the simultaneous occurrence of more than one of these illnesses is impossible” means that $\displaystyle P(H) = P\left( {H \cap I_1 } \right) + P\left( {H \cap I_2 } \right) + P\left( {H \cap I_3 } \right).$
From the given you can find each of these: $\displaystyle P\left( {H \cap I_1 } \right) = P\left( {H|I_1 } \right)P\left( {I_1 } \right)$