# (none)

Les Willis WILLISL at EM.AGR.CA
Tue Apr 30 12:09:02 EST 1996

```

I believe you can use the probability equation from Sambrook:

N = ln(1-P)/ln(1-f),  where P is the desired probability, f is the fractional
proportion of the genome in a single recombinant, and N is the
necessary number of recombinants (Clarke and Carbon, 1976).

eg. for a 99% probability of a 17 kb fragment in a mammalian genome of
3 X 10 E09 bp:

N=                          ln(1- 0.99)                        =  8.1 X 10 E05
--------------------
ln( 1- [1.7 X 10 E04/ 3 X10 E09])

I've been trying to figure out something that should be simple but I seem  to be stuck.
Perhaps someone can help. I need to figure out what the  probability of NOT finding a
fragment which is r bp long in a sequence  which is n bp long. I can do this for
individual cases but I need a  purely symbolic solution which would cover all cases.
This is all  assuming a completely random sequence.  If anyone happens to know off
the top of their heads what the solution is  it would be grotesquely appreciated.
Thanks,