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
 A: 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}
A: The possibility of a woman getting ill $1:350$
The possibility that ultrasound is inaccurate : $1:4$
You need both to occur, so the chances of that happening is $1:350*4 = 1 :1400$, as they are independent events.
A: 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 $\frac{1}{4}\times\frac{1}{350}=\frac{1}{1400}$.
A: Unfortunately, the question is imprecise, and with it explanations in answers.
This is a classic screening test question in a medical setting, for which Bayes' Theorem provides an answer--when the necessary information is provided. 
The required answer for the question at hand is the 'predictive power of a negative test'
$\delta = P(No\, Disease | Negative\, test).$
To answer, we need to know the prevalence of the disease in the population, $\psi = P(Disease)$.  Here we are given $\psi = 1/350.$
We also need to know both the sensitivity of the test $\eta = P(Positive\, test|Disease)$
and the specificity of the test $\theta = P(Negative\, test|No\, disease).$
Saying the test has a certain "accuracy" is imprecise, because both $\eta$ and $\theta$ reflect the usefulness of the test. It is unclear whether the
75% figure given in the question is meant to apply to sensitivity or to specificity.
When the required information is known, it follows that the proportion of the population receiving positive tests is 
$$\tau = P(Positive\, test)= P(Pos\;and\; Dis) + P(Pos\;and\;No\,dis) = \psi \eta + (1-\psi)(1-\theta).$$
Then a simple application of
Bayes' Theorem gives the probability $\delta = (1-\psi) \theta/(1-\tau).$
In your question, you might suppose the "accuracy" claim means $\eta = \theta = .75$. Then $\tau = (1/350)(.75) + (349/350)(.25) =  0.2514$ and
$\delta = (349/350)(.75)/(1-.2514) \approx  0.9990.$ So there would about one chance in a thousand a person with a negative test has the disease.
Or you might suppose "accuracy" means $\eta = .75$ and guess that $\theta \approx 1$--the latter assuming (with more hope than realism) that the ultrasound is incapable of giving a false alarm. Then $\tau \approx \psi\eta = (1/350)(.75) =  0.002143$ and $\delta \approx (349/350)/(1-0.002143) = 0.9993.$
And there would be about 7 chances in 10,0000 of having a disease.
Note: The usual textbook question for screening tests is to ask for the predictive power of a positive test,
$\delta = P(Disease|Positive\, test) = \psi\eta/\tau.$ When $\psi$ is very
small, $\delta$ can be surprisingly small, even when $\eta$ and $\theta$ are both around 95%. 
Unfortunately, when sloppy language about the "accuracy" of a screening test is used, it can be taken by different individuals to mean high $\eta$, $\theta,$ or $\delta.$ This kind of imprecision can make it almost impossible to have a rational discussion about the public health policy or homeland security policy of various kinds of screening tests.
For an excellent general discussion of various uses of screening tests, see Gastwirth, Statistical Science (1987). One introductory technical discussion is Ch.5 of Suess and Trumbo, Springer (2010). 
