words = ['rato', 'roeu', 'rolha', 'rainha', 'rei', 'russia']
pref = ['ro', 'ra', 'r']

contTotal = 0
for p in pref:
    cont = 0
    for w in words:
        if w.startswith(p):
            cont+=1
    contTotal += cont
    print p + ' aparece ' + str(cont) + ' vezes nas palavras como prefixo'
print 'O numero total de vezes é ' + str(contTotal)

Abominable functional One-liner with sum and map :

sum(map(lambda x: 1 if x.startswith(tuple(pref)) else 0, words))

Abominable one-liner with reduce :

reduce(lambda x, y: x + 1 if y.startswith(tuple(pref)) else x, words, 0)

:)

Update:

As per requirements of the OP, even more abominable one-liner of snows:

map(lambda p: reduce(lambda c, w: c + 1 if w.startswith(p) else c, words, 0), pref)

Less forced example:

def countPrefix(words, prefix):
    return len([1 for w in words if w.startswith(prefix)]) 

[countPrefix(words, p) for p in pref]   

Result:

[2, 2, 6]

I could do it this way:

>>> palavras = ['rato', 'roeu', 'rolha', 'rainha', 'rei', 'russia']
>>> prefixos = ['ro', 'ra', 'r']
>>> len(filter(None, [p if p.startswith(tuple(prefixos)) else None for p in palavras]))
6

* In this case to know how many words had one of the occurrences of the list.

Assuming that there is no "hierarchy" between prefixes (eg, every word that begins with ro also starts with r ), a simple and straightforward way is using itertools.product . It will combine each element of the first list with each element of the second list. So just filter those that the second is prefixed from the first, and count:

>>> import itertools
>>> palavras = ['rato', 'roeu', 'rolha', 'rainha', 'rei', 'russia']
>>> prefixos = ['ro', 'ra', 'r']
>>> len([palavra for palavra,prefixo in itertools.product(palavras, prefixos) if palavra.startswith(prefixo)])
10

Note that the first 4 words have been counted twice (since they begin with both ro / ra and r ). For a solution that only counts each word once, see for example the responses from Orion and Anthony Accioly (the Dherik does both, and still counts occurrences by prefix).