Chapter 12.EA, Problem 1EA

Find $P\left(H_2|C_1\right)$.

To determine

To find:

The value of $P\left(H_2|C_1\right)$.

Answer to Problem 1EA

Solution:

The required probability is $P\left(H_2|C_1\right)=0.08$.

Explanation of Solution

Given:

The probability of different state of health is given below:

	State of Health, $H$	$P\left(H\right)$
$H_1$	Patient has a normal heart rate	$0.8$
$H_2$	Patient has minor heart irregularities	$0.15$
$H_3$	Patient has a severe heart condition	$0.05$

The cardiogram produces two results:

$C_1=$ Stethoscope and cardiogram shows normal heart rate.

$C_2=$ Stethoscope shows normal heart rate and cardiogram shows minor irregularities.

$\begin{array}{l}P\left(C_1|H_1\right)=0.9\\ P\left(C_1|H_2\right)=0.4\\ P\left(C_1|H_3\right)=0.1\end{array}$

Approach:

Bayes’ theorem shows that $P\left(A|B\right)=\frac{P\left(B|A\right)P\left(A\right)}{P\left(B\right)}$.

Here, $P\left(A|B\right)$ is probability of occurring $A$ given $B$ is true,

$P\left(B|A\right)$ is probability of occurring $B$ given $A$ is true,

$P\left(A\right)$ and $P\left(B\right)$ are the probabilities of occurring $A$ and $B$ respectively.

Calculation:

Use the Bayes’ Theorem to calculate $P\left(H_2|C_1\right)$.

$\begin{array}{rll}P\left(H_2|C_1\right)& =& \frac{P\left(C_1|H_2\right)P\left(H_2\right)}{P\left(C_1|H_1\right)P\left(H_1\right)+P\left(C_1|H_2\right)P\left(H_2\right)+P\left(C_1|H_3\right)P\left(H_3\right)}\\ & =& \frac{0.4\left(0.15\right)}{0.9\left(0.8\right)+0.4\left(0.15\right)+0.1\left(0.05\right)}\\ & =& \frac{12}{157}\\ & \approx & 0.08\end{array}$