# Math Help - conditional probability proof

1. ## conditional probability proof

Q: Show that $P(T|D^{c})=1-P(T^{c}|D^{c})$

A:This is my path...

$P(T|D^{c})=\frac{P(T\cap\\D^{c})}{P(D^{c})}=$ $
\frac{P(T/D)}{P(D^{c})}=\frac{P(T)-P(T\cap\\D)}{P(D^{c})}=
\frac{P(T)-P(D)P(T|D)}{P(D^{c})}=$
$\frac{[P(T)-P(D)]P(T|D)}{P(D^{c})}=$

2. Here is what I would do.
$P(T|D^c ) = \frac{{P(T \cap D^c )}}
{{P(D^c )}} = \frac{{P(D^c ) - P(T^c \cap D^c )}}
{{P(D^c )}} = 1 - P(T^c |D^c )$