There are a lot of studies on the probability of accidents in a nuclear power plant. As far as I understand they use methods of risk analysis to calculate the failure probability of the nuclear reactor.
Here I tried a very simple empirical approach: We know the number of nuclear power reactors in the world, we know (probably) the number of severe accidents up to now, so we can calculate the empirical failure probability of a single reactor per year. Thus we are able to calculate the probability that no reactor in the world, in UK or in another country, will have an accident within the next 5, 10 or 20 years.Or that at least on reactor will fail severely. This can be done by using the Poisson distribution.
Up to now there are at least 4 reactor accidents on INES scale 5 or more. Chernobyl (1986) is the only one on level 7, Three Miles Island(1979), Windscale (1957) are on level 5. Also the present Fukushima accident (or accidents?) is level 5, at least at the moment (27.03.2011). On level 5 there are some more accidents and on level 6 is only one, but they were in other nuclear facilities, not in power reactors. One could argue that Windscale was not a civil but a military reactor, but then in Fukushima there is probably more than one reactor involved. So the number of 4 severe accidents seems quite reasonable.
The number of nuclear reactors worldwide increased drastically from 1955 until 1988, from which date the number is nearly constant. Up to the Fukushima accident there were 443 reactors operating worldwide.
By a simple graphical piecewise interpolation of the number of reactors per year a total of 15.000 reactoryears can be estimated. This crude number should be sufficient for the present purpose.
So the probalilty for one severe accident per reactoryear (ry)is
q=4acc/15.000ry
If there are N reactors in operation, the Poisson distribution gives the probability for x severe accidents within the next y years. In order to apply the Poisson distribution the expected mean number of accidents m within this time has to be estimated:
m=q*N*y
Then the probality to have x accidents when we expect a mean value of m accidents is given by
p(x)=(m^x/x!)*exp(-m)
Thus the probality for no accident is (x=0)
P(0)=exp(-m)
and the probality for at least one accident is
p(x≥1)=1-exp(-m)
Regarding the worldwide situation for the next 20 years, the number of reactors is 443, we expect an average number of severe accidents
m=(4acc/15.000ry)*443r*20y=2.34acc
so 2.34 accidents within any period of 20 years somewhere in the world. The probability for one or more severe accidents worldwide is
p(x≥1)=1-exp(-2.34)=90.36%
How is the situation for a single country? We simply have to count the number of reactors within this country and calculate the respective reactoryears.
                                      World        UK               US            D
reactors                       443           19              104           17
reactoryears             8860        380            2080         340
mean # acc                 2.34         0.100       0.549         0.089
p(≥1                              90.36%    9.55%      42.26%       8.59%
On the average more than 2 accidents are expected worldwide, the probality for at least one accident ist 90% worldwide, more than 9% for the UK and more than 40% for the US.
Do you think these estimations are reasonable? Do you think a 9% probability for a Chernobyl or Fukushima accident in the UK within the next 20 years is acceptable?

I am looking forward to your comments.