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At a hospital's emergency room, patients are classified and 20% of them are critical, 30% are serious, and 50% are stable. Of the critical ones, 30% die; of the serious, 10% die; and of the stable, 1% die. Given that a patient dies, what is the conditional probability that the patient was classified as critical?

Wouldn't it be P(Crit.| Die) = (.2 x .3 / .3) ? I don't know what i'm doing wrong. The answer is 0.632

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Let $C$ be the event a randomly chosen patient is critical, and $D$ be the event that the event that a randomly chosen patient dies. We want $\Pr(C|D)$. By the usual formula for conditional probability, we have $$\Pr(C|D)=\frac{\Pr(C\cap D)}{\Pr(D)}.\tag{1}$$ We want to find the probabilities on the right of (1).

The probability $\Pr(C\cap D)$ is, as you calculated, $(0.2)(0.3)$.

Now we need $\Pr(D)$. The event $D$ can happen in $3$ ways: (i) critical and dies or (ii) serious and dies or (iii) stable and dies. Calculate these probabilities and add up. You have already found the probability of (i).

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Let $A=$ the event that the patient is in critical condition, $B=$ the event that the patient is in serious condition, and $C=$ the event that the patient is in stable condition. Let $D=$ the event that the patient dies. The we know $P(A)=0.2, P(B)=0.3, P(C)=0.5, P(D|A)=0.3, P(D|B)=0.1, P(D|C)=0.01$. Then $$P(A|D)=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)}$$ For further explanation read Baye's Theorem.

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The probability of A given B is equal to P(A|B) = $\frac {P(B | A)\, P(A)}{P(B)}$. In your problem, B = "Given that a patient dies", and A = "the probability that the patient was classified as critical". This is known as Baye's Theorem. I suggest you read up on it as it is very important in probability.

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