# Combinatorics in Probability

• Mar 3rd 2011, 09:29 AM
ragnar
Combinatorics in Probability
I'm looking at someone's solution to a probability problem, which is: On a true/false test of ten questions what is the probability of getting exactly 7 right? My answer makes sense to me. We want the distinguishable permutations of TTTTTTTFFF over 2^10. The distinguishable permutations are 10!/7!3!.

Her answer computes to the same, but she does it by 4(10c7). I see that these are basically the same but my problem is how should one see that the choice function applies here? I interpret the choice function to mean the number of possible subsets of a set of ten objects, each subset having seven objects. How does that describe the original question? And what if you didn't have just true/false? Suppose the question were about picking balls out of an urn where three are red, three white, and three blue (makes you want to sing the national anthem, right?). Then the number of distinguishable permutations is 9!/3!3!3!. How can you get that from use of the choice function?
• Mar 3rd 2011, 10:21 AM
Plato
Quote:

Originally Posted by ragnar
I'm looking at someone's solution to a probability problem, which is: On a true/false test of ten questions what is the probability of getting exactly 7 right? My answer makes sense to me. We want the distinguishable permutations of TTTTTTTFFF over 2^10. The distinguishable permutations are 10!/7!3!.

Quote:

Originally Posted by ragnar
Her answer computes to the same, but she does it by 4(10c7).

Unless you have mis-quoted her answer, there is no way that can be correct.

Quote:

Originally Posted by ragnar
Suppose the question were about picking balls out of an urn where three are red, three white, and three blue (makes you want to sing the national anthem, right?). Then the number of distinguishable permutations is 9!/3!3!3!. How can you get that from use of the choice function?

The first answer is correct. This problem does not rely on a choice function.
• Mar 3rd 2011, 10:23 AM
ragnar
Er, that 4 is in there by accident. But 10c7 = 10!/3!7! which is the same as my result.