probability of illness There is a 1 in 350 chance of having an illness for a woman.  

An ultrasound is 75% accurate in establishing whether the illness is present. 

If the ultrasound is negative (i.e. suggests that the woman does not have the illness) what is the new probability of the woman actually having the illness?  

I just got some results and am not sure how to connect the two probabilities.  I am as interested in filling my knowledge gap about how to calculate the probability as I am in the actual answer.  

Thanks,

Ivan
Define events

$D$: the woman has this disease

$A$: the test result is positive

We have $\ \mathbb{P}(D)=1/350$, $\ \mathbb{P}(A|D)=\mathbb{P}(A^C|D^C)=0.75$

The problem is asking: what is the probability $\mathbb{P}(D|A^C)$ ?

\begin{align}
\mathbb{P}(D|A^C) =\dfrac{\mathbb{P}(D\cap A^C)}{\mathbb{P}(A^C)}
=\dfrac{\mathbb{P}(D\cap A^C)}{\mathbb{P}(D\cap A^C)+\mathbb{P}(D^C\cap A^C)},
\end{align}
where we have
\begin{align}
\mathbb{P}(D\cap A^C) = \mathbb{P}(D) - \mathbb{P}(D\cap A)
=\mathbb{P}(D)-\mathbb{P}(A|D)\,\mathbb{P}(D)
=\mathbb{P}(D)\big[1-\mathbb{P}(A|D)\big]
=\dfrac{1}{4}\times\dfrac{1}{350}
\end{align}
and
\begin{align}
\mathbb{P}(D^C\cap A^C)=\mathbb{P}(A^C|D^C)\,\mathbb{P}(D^C)
=\dfrac{3}{4}\times\dfrac{349}{350}
\end{align}
Thus
\begin{align}
\mathbb{P}(D|A^C)=\dfrac{1}{1+3\times349}
\end{align}
There's a $\frac{1}{4}$ chance the test is incorrect.

There's a $\frac{1}{350}$ chance the woman has the illness.

Assuming that these are independent events (there's no evidence in your question that they shouldn't be), so the probability both occurs is $\frac{1}{4}\times\frac{1}{350}=\frac{1}{1400}$.