In my intro stats class, we are learning about Bayes' theorem, and I am a bit confused. In an example problem, we are given that for a medical patient:

$P(\text{death} | \text{2 organ failures}) = 0.85$

I am trying to calculate what

$P(\text{survival} | \text{2 organ failures})$ should be. I am not sure how to apply the total probability here, so should it be just 1 - 0.85 = 0.15?


  • 1
    $\begingroup$ Assuming there isn't any other possibilities that can happen, then yes! The probability of survival given two organ failures is $1-.85 = .15$ $\endgroup$
    – WaveX
    Sep 15, 2019 at 0:15
  • $\begingroup$ If it helps alleviate your confusion, this isn't Bayes' theorem. This is just conditional probability. Bayes' theorem is what you would use if you wanted to calculate P(2 organ failures | death). $\endgroup$
    – user694818
    Sep 15, 2019 at 8:56

1 Answer 1


Let $D$ denote the event of death and $O$ the event associated to the failure of two organs. Since $O = O\cap\Omega = O\cap(D\cup\overline{D}) = (O\cap D)\cup(O\cup\overline{D})$ and $(O\cap D)\cap(O\cap\overline{D}) = \varnothing$, where $\Omega$ is the sample space, one has \begin{align*} &\textbf{P}(O) = \textbf{P}(O\cap D) + \textbf{P}(O\cap\overline{D}) \Longrightarrow \textbf{P}(O\cap\overline{D}) = \textbf{P}(O) - \textbf{P}(O\cap D) \Longrightarrow\\\\ & \textbf{P}(\overline{D}| O) = \frac{\textbf{P}(\overline{D}\cap O)}{\textbf{P}(O)} = \frac{\textbf{P}(O) - \textbf{P}(O\cap D)}{\textbf{P}(O)} = 1 - \frac{\textbf{P}(O\cap D)}{\textbf{P}(O)} = 1 - \textbf{P}(D|O) \end{align*}

$\therefore \textbf{P}(\overline{D}|O) = 1 - 0.85 = 0.15$

Another possible (and direct) approach consists in observing that the conditional probability is a probability measure itself. Therefore we have that $\textbf{P}(D|O) = 1 - \textbf{P}(\overline{D}|O)$. Hope this helps.


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