What is the hybridisation of the carbons in the allyl radical? The example I am talking about is $\ce{CH2CHCH2^.}$.

To find the steric number and hence the hybridization, we use steric number  = No. of lone pairs + number of sigma bonds. In this example, the last C atom has only one electron. Should we count it as a lone pair? 

My teacher says that that electron will get delocalised so its hybridization will be $\mathrm{sp^2}$ and not $\mathrm{sp^3}$ (which I think is true). I didn't understand the delocalisation stuff. Is it true that here, there will be delocalisation and if not then what is happening here? If delocalisation is occurring here then please explain it too! Also please tell me in which cases is there delocalisation of electron(s)?

My teacher also said that had it been $\ce{CH3CH2CH2^.}$, then there would be no delocalisation of electrons. Why does this happen?
Radicals are electron-deficient compounds. You should not treat a radical as a lone pair, as they are pretty different.

A good rule of thumb is to assume an orbital that contains the highest possible p-contribution for the single electron. The more s-contribution is mixed in, the ‘closer’ the single electron gets to the nucleus which is less favourable than having an electron pair in a similar orbital. A pure p-orbital fulfils the condition perfectly by having no s-contribution whatsoever. Thus, you should assume radicals to be in p-orbitals whenever possible.

This is similar to carbocations which have an orbital fully devoid of electrons. However, this empty orbital must always be a p-type orbital for electronic reasons. Radicals have the possibility of undergoing hybridisation and e.g. ending up in an $\mathrm{sp}^n$ orbital (examples: radicals on $\mathrm{sp^3}$ hybridised bridgehead atoms; phenyl radical).

Since we just established that the radical will occupy a p-type orbital and since there is a &#x3c0; bond nearby, we can delocalise. In resonance structure terms, this can be visualised in the following way:

$$\ce{H2C=CH-CH2^. &lt;-&gt; H2C^.-CH=CH2}$$

The radical is delocalised across two carbon atoms making the overall system more stable than if it weren’t. In molecular orbital terms, this is basically a stabilisation of the double bond at the expense of the radical-containing orbital which benefits the overall system. The delocalisation can, of course, only occur if there is a &#x3c0; bond nearby; the n-propyl radical does not feature one.