I have been given a question to answer and am unsure if I am making the right calculations. The question is:

**A 4-cutter restriction enzyme such as Hae lll recognises a particular sequence of four base pairs. The probability that any given set of 4 bases is the Hae lll recognition site is 1/4 x 1/4 x 1/4 x 1/4. So, what would you expect to be the average fragment size when human DNA is cut with Hae lll?**
So far, I have made the following calculations:

There are 6,000,000,000 base pairs in human DNA and 1/4^4 is 0.00390625.

So, 0.00390625 x 6,000,000,000= 23437500

Am I on the right lines? If anyone could give me some help with this question it would be great.Thakyou in advance.