conditional probability with complements, false negative

Question

given $P(T|D) = .95, P(T|D^c) = .1 , P(D) = .01$ I found $P(D|T) = .904762$ and I want to find $P(D|T^c)$ but I am unsure how to proceed. Thanks

Answer 1

To begin with, start by noticing that \begin{align*} \textbf{P}(D|T^{c}) = \frac{\textbf{P}(D\cap T^{c})}{\textbf{P}(T^{c})} = \frac{\textbf{P}(D) - \textbf{P}(D\cap T)}{1 - \textbf{P}(T)} \end{align*} This is because \begin{align*} \textbf{P}(D) = \textbf{P}(D\cap\Omega) = \textbf{P}(D\cap(T\cup T^{c})) = \textbf{P}(D\cap T) + \textbf{P}(D\cap T^{c}) \end{align*} Based on the given data, one has \begin{align*} \begin{cases} \displaystyle\textbf{P}(T|D) = \frac{\textbf{P}(T\cap D)}{\textbf{P}(D)} = 0.95\\\\ \displaystyle\textbf{P}(T|D^{c}) = \frac{\textbf{P}(T\cap D^{c})}{\textbf{P}(D^{c})} = \frac{\textbf{P}(T) - \textbf{P}(T\cap D)}{1 - \textbf{P}(D)} = 0.1 \end{cases} \end{align*}

Since you know the value $\textbf{P}(D) = 0.01$, you can solve the system of equations in terms of $\textbf{P}(T)$ and $\textbf{P}(D\cap T)$. To be precise, this is the system of equations \begin{align*} \begin{cases} \displaystyle\textbf{P}(D\cap T) = 0.95\times\textbf{P}(D) = 0.0095\\\\ \displaystyle\textbf{P}(T) - \textbf{P}(T\cap D) = 0.099 \end{cases} \end{align*}

Answer 2

given $P(T|D) = .95, P(T|D^c) = .1 , P(D) = .01$

With this, it is straightforward.

Just do the same as you did to find that $P(D\mid T)=1/12$, remembering that $P(T^c\mid D)=1-P(T\mid D)$.

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$\begin{align}P(T) &=P(T\mid D)\cdot P(D)+P(T\mid D^c)\cdot (1-P(D)) \\[2ex] P(D\mid T) &=\dfrac{P(D)\cdot P(T\mid D)}{P(T)}&&=\tfrac 1{12} ~\checkmark \\[2ex]P(D\mid T^c)&= \dfrac{P(D)\cdot(1-P(T\mid D))}{1-P(T)}\end{align}$