Let's say each kid flips a biased coin to determine whether or not they have cancer. If we assume that the probability of heads (cancer) is 1.6/100,000, we can find the distribution of cancer counts we'd expect using a binomial distribution.
R code, we can find the distribution with the
dbinom(x = 0:5, size = 38000, prob = 1.6/100000)
x is the number of cases (
0:5 means we're looking at the probability of 0 cases, 1 case, etc. up to 5). Size is the number of kids, and prob is the baseline probability you cited.
After cleaning up the output slightly, we get a table like this:
So you'd expect to find 3 cases out of 38,000 children only 2% of the time under this model--and you'd almost never find more than that.
In short, (assuming the figures are comparable), it does seem on the high side, and might be worth investigating further. But you wouldn't necessarily need to invoke any special factors beyond random chance to explain the difference.
Edited to add: Per EpiGrad's comment, I added an image that shows how these probabilities could change if we were uncertain about the baseline probability of 1.6 cases per 100k. The red points are the values I listed above, and the cloud of points represent what we'd expect if the
For this example, I sampled baselines from a beta distribution using
rbeta(1000, 1.6, 100000 - 1.6), which has a mean of 1.6 cases per 100k and some spread on either side but doesn't drop below 0. The amount of spread may or may not be reasonable, depending on what assumptions you'd like to make. My gut feeling is that I included more variation than I should have, but who knows.
As you can see from the plot, if the British figures substantially underestimated the "real" pre-Fukushima rate of cancer incidence in Japan, we might expect to see 3 cases per 38k as often as 20% of the time. Whether you think that's likely depends on other information outside the scope of this problem, including whether I included an appropriate amount of uncertainty from the British estimate.