# Binomial Probability of Success Given Probabilities of Individual Words in a Phrase?

If I have aggregated statistics on individual words of a phrase, is there a way to compute the overall probability of a phrase having a "success" vs "not success" (binomial)?

Example

Phrase: 8 inch wood lap siding

I have individual statistics on each word in this phrase.

1. When 8 is in the phrase, "success" is 1/100 or 1%
2. When inch is in the phrase, "success" is 1/200 or 0.5%
3. When wood is in the phrase, "success" is 1/80 or 1.25%
4. When lap is in the phrase, "success" is 1/125 or 0.8%
5. When siding is in the phrase, "success" is 1/25 or 4.0%

What I'm trying to determine is if there is some mathematical way to compute the probability that a phrase will have a "success" if I know the individual probabilities of each word.

I don't know if this approach conceptually makes sense. Some words have high collocation with others so I don't know if this approach would capture that. Plus, some words appear very frequently (have heavier weights) and others don't.

• Do you consider those probabilities to be independent? Do you have only information about marginal probabilities? If you have full dataset, that why not just simply use logistic regression?
– Tim
Commented Nov 23, 2015 at 18:13
• @Tim Not sure how to answer that. If I have 5 keywords, each keyword has a number of trials and successes. I split each keyword by space into words, then group by each word and sum trials and successes. A word would only appear in a keyword a max of 1 time but never twice or more. Ex: this is a keyword with 1/100 overall probability of "success" splits to this is a keyword but an important thing to note is that each word is assigned the overall probability of 1/100 equally which artificially creates more trials and "successes" when summed. Would logistic regression be a solution? Commented Nov 23, 2015 at 22:41