Talks about Covid-19 are all around. I was wondering what is the probability that someone who tests +ve with the virus would die. It is scary to ask this but it is maybe useful to know. Here are the (made up) events and probabilities.
Consider these events: Getting infected with the virus = $I$, Getting detected (by medical equipment)= $D$, Death/mortality = $M$
Given, $P(I) = 0.01, P(D\mid I)=0.98, P(M\mid I)=0.03$, We need to find $P(M\mid D)$
Here is my solution:
$$ P(I\mid D) = \frac{P(D\mid I)*P(I)}{P(D)} \\ P(D\mid I) = 0.98\\ P(I) = 0.01 \\ P(D) = P(D\mid I)*P(I) + P(D\mid I^\complement)*P(I^\complement) = 0.98*0.01 + 0.02*0.99 = 0.0296\\ P(I\mid D) = 0.98*0.01/0.0296 = 0.331\\ P(M\mid D) = P(M\mid I)*P(I\mid D)\\ P(M\mid D) = 0.03*0.331 = 0.00993 $$
Most examples in the book ends at $P(I\mid D)$. I have doubt about $P(M\mid D) = P(M\mid I)*P(I\mid D)$. Is it correct? If not what is the right way of solving this problem.
(I wish everyone gets recovered from covid-19 soon).
Edit-1 (Here is the updated question and data based on @joriki's answer)
Consider the events: Getting infected with the virus = $I$, Getting detected (by medical equipment)= $D$, Death/mortality = $M$
Given, $P(I) = 0.01, P(D\mid I)=0.98, P(M\mid I,D)=0.03, P(D\mid I^\complement)=0.05$, find $P(M\mid D)$
Here is the NEW solution:
$$ P(I\mid D) = \frac{P(D\mid I)*P(I)}{P(D)} $$ $$P(D\mid I) = 0.98$$ $$P(I) = 0.01 \\ P(D) = P(D\mid I)*P(I) + P(D\mid I^\complement)*P(I^\complement) = 0.98*0.01 + 0.05*0.99 = 0.0593$$ $$P(I\mid D) = 0.98*0.01/0.0593 = 0.165$$ $$P(M\mid D) = P(M\mid I,D)*P(I\mid D) + P(M\mid I^\complement,D)*P(I^\complement\mid D)$$ Also, $P(M\mid I^\complement,D) = 0$ $$P(M\mid D) = P(M\mid I,D)*P(I\mid D) = 0.03*0.165 = 0.00495$$
The information that you’re assuming is not enough to determine the probability you’re seeking.
You seem to have used $P(D\mid I^\complement)\stackrel?=1-P(D\mid I)$. That's not true in general. You can treat $P(D\mid I)$ like a probability for $D$, so $P(D^\complement\mid I)=1-P(D\mid I)$, but you can't do the same thing with the condition: The probability that I eat cereal tomorrow if it rains isn’t the complement of the probability that I eat cereal tomorrow if it doesn’t rain; it’s the complement of the probability that I don’t eat cereal tomorrow if it rains.
$P(M\mid D)\stackrel?=P(M\mid I)P(I\mid D)$ is also not true in general. What’s true in general is
$$ P(M\mid D)=P(M\mid I,D)P(I\mid D)+P(M\mid I^\complement,D)P(I^\complement\mid D)\;. $$
In the present case, though, you could argue that $P(M\mid I^\complement,D)=0$, since you’re certain not to die of the disease if you’re not infected; that would yield
$$ P(M\mid D)=P(M\mid I,D)P(I\mid D)\;. $$
This is equal to what you used if $P(M\mid I,D)=P(M\mid I)$. That’s a rather unplausible assumption, though; presumably mortality and detection are both correlated to the severity of the symptoms. More plausibly, you could say that what you can actually estimate from the data isn’t $P(M\mid I)$ but $P(M\mid I,D)$, so you might want to change your assumption to $P(M\mid I,D)=0.03$.
That still leaves the problem that you need $P(D\mid I^\complement)$, which is independent of the data you’ve assumed so far.