Interpretation of Bayes Formula Question

Question

This is a simple application of Bayes Theorem, I'm just confused about the labeling. If i let T = test outcome is positive, D = patient has disease, then, I am trying to find

$$ P ( D | \bar{T} ) = \frac{P(D,\bar{T})}{P(\bar{T})}\\ =\frac{P(\bar{T}|D)P(D)}{P(\bar{T}|D)P(D)+P(\bar{T}|\bar{D})P(\bar{D})}\\ =\frac{0.25 \cdot 0.25}{0.25 \cdot 0.25 + 0.75 \cdot 0.75} =0.1 $$

Is this the correct interpretation?

Answer 1

Your interpretation of the numbers appears correct. Your concern was over whether you correctly interpreted what $P(\overline{T}|D)$ was. You may use and/or memorize the following result:

$P(A|B) = 1 - P(\overline{A}|B)$

Proof:
$$\begin{align} P(A|B) & = \frac{P(A\cap B)}{P(B)} & \text{by definition of conditional probability}\\ & = \frac{P(A\cap B) + P(\overline{A}\cap B) - P(\overline{A}\cap B)}{P(B)} & \text{by adding "zero"}\\ & = \frac{P((A\cap B)\cup(\overline{A}\cap B)) - P(\overline{A}\cap B)}{P(B)}& \text{since}~A\cap B~\text{is disjoint from}~\overline{A}\cap B\\ &=\frac{P((A\cup \overline{A})\cap B) - P(\overline{A}\cap B)}{P(B)} &\text{by distributivity}\\ &=\frac{P(S\cap B) - P(\overline{A}\cap B)}{P(B)} & \text{by definition of compliments}\\ &=\frac{P(B) - P(\overline{A}\cap B)}{P(B)} & \text{by definition of sample space}\\ &=\frac{P(B)}{P(B)} - \frac{P(\overline{A}\cap B)}{P(B)}&\text{by algebra}\\ &=1 - P(\overline{A}|B) & \text{by definition of conditional probability}\end{align}$$