## Given¶

The number of restriction sites in a fragment of length \(t\) are distributed as:

$$
P(N_t=k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}
$$

Find the distribution of length of restriction fragments

## Solution¶

Intuition: If I have a restriction site at some location \(z\) and want it to be atleast \(x\) base pairs long, i.e. I have a fragment starting at \(z\) and going atleast till \(z+x\), I need to ensure I do not encounter any other restriction site in between, which is also equivalent to \(P(N_x=0) = e^{-\lambda x}\)

Thus, the probability that the restriction site is atleast \(x\) bp is given by:

$$
P(X > x) = e^{-\lambda x}
$$

and hence,

$$
P(X < x) = 1-e^{-\lambda x} \implies P(X=x) = \lambda e^{-\lambda x}
$$

Thus, the size of restriction fragments follows an exponential distribution.

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