Algorithm Suggestion for Scoring System - weighted solution
I don't think this question is related at all with approximation algorithms, or theoretical CS.

Please take the following as some free, non-exhaustive, thoughts on your question.

It seems to me that what you want is just the probability that a given sequence of words contains only valid words.

Let me assume that each word is chosen independently (which it isn't true).
Suppose we have $k$ words, call $\sigma_i$ the i-th of such words.
Call $A_i$ the event "$\sigma_i$ is a valid word" and $A$ the event "all words are valid".

Under this assumptions you have:
$$
P(A) = \prod_{i=1}^{k}P(A_i)
$$

so you only need to assign a probability to each word (and you've already given such a distribution in you question).
However you may want to apply some smoothing: if you set $P(A_i)=0$ for some $A_i$ then also $P(A)$ will be $0$ (if you are sure that a word is not valid, then then the whole phrase cannot be valid).

A bayesian approach only requires conditional independency. For example you can compare the following two quantities (and choose the highest):

$$
P(A | \sigma_1, \dots, \sigma_k) \propto  P(\sigma_1, \dots, \sigma_k | A) P(A) = \prod_{i=1}^{k}P(\sigma_i|A) P(A)
$$
$$
P(A^C | \sigma_1, \dots, \sigma_k) \propto  P(\sigma_1, \dots, \sigma_k | A^C) P(A^C) = \prod_{i=1}^{k}P(\sigma_i|A^C) (1 - P(A))
$$

where $P(\sigma_i|A)$ is the probability that the word $\sigma_i$ appears in a valid phrase, $P(\sigma_i|A^C)$ is the probability that $\sigma_i$ appears in an invalid phrase and and P(A) is the probability that a valid phrase shows up.

In the above we are just differentiating "valid" from "invalid" phrases. If you want to account for "completely valid", "completely invalid" and "mixed up" phrases you can do something similar.

You'll need to assign three probabilities to each word $\sigma_i$: $P(\sigma_i|valid)$, $P(\sigma_i|invalid)$, $P(\sigma_i|mixed)$ and three "a priori" probabilities, for example:


$P(valid) = 0.55$  (most likely)
$P(invalid) = 0.40$ (likely)
$P(mixed) = 0.05$ (not likely)