Question about Autosomal Recessive Alleles
If really cystic fibrose is expressed only in homozygotes, then here are some possible explanations:


Mutation

very improbable

Autofecundation

very improbable

False Negative when testing the father

Because of the test False Negative Rate. @kmm made a very interesting comment below. He said that according to this article 75% of the mutations coding for disease correspond to one pattern (3 base pair deletion). The other 25% of the mutations coding for disease might not be detected in the standard test. If so, it is very likely that a the test conducting on the father was a false negative. It would be worth further investigate the test that are usually performed to confirm this.
Because of human error. The laboratory operator might accidentally switch two samples or accidentally introduce stranger DNA before PCR (assuming the test is performed through sequencing)
I have no idea concerning the probability for such thing to occur! 

trisomy

The assumes that a trisomic that have two recessive allele coding for disease and on dominant allele coding for health yield to a individual expressing the disease!
The gene for cystic fibrose is on chromosom 7. Trisomy 7 is extremely rare if I'm not mistaken.

The father is not the father!

According to different studies between 5% and 20% of the people are not fathered by the man they think is their dad.
According to this source the frequency of the disease is 1/2500 in UK. Assuming Hardy-Weinberg equilibrium the frequency of the allele coding for disease should be $p=0.02$.
The probability that the baby comes from another man and that the baby express the disease is $$a \cdot \left(\frac{2\cdot p\cdot (1-p)}{4} + \frac{p^2}{2}\right)$$, where $a$ is the probability that the baby was fathered by another man (let's say 10%) and $p$ is the frequency of the allele coding for the disease (0.02). The division by 2 and 4 are due to the probability that the (real) father and the mother transmit both the alleles coding for the disease.  This probability equals $$0.1 \cdot \left(\frac{2\cdot 0.2\cdot (1-0.2)}{4} + \frac{0.2^2}{2}\right) = 0.01$$ This seems to be from far the most likely explanation! Note: this assumes that the mother is heterozygote (does not express the disease) otherwise, the probability would be 0.02.