N of possible answers on a filtered test with different numbers of passing options

  • #1

Main Question or Discussion Point

Hopefully I'm posting this to the appropriate section. I am trying to figure out a formula for describing the number of possible answer "paths" that may be taken by an individual who is administered a filtered test format. In the filtered test format, the individual must select a passing response to item j in order to proceed to item j + 1. If the individual selects a failing option, the test is over. My question is, given that each of the items on the test can have different numbers of passing and failing options (e.g., item j might have two passing options and three failing options, item j + 1 might have one passing option and two failing options), how can I even begin to figure out how to put this down mathematically on paper? My math background is rather weak, so I'm not really even sure where to start. Any help would be much appreciated!
 

Answers and Replies

  • #2
34,387
10,475


You can calculate it step by step: For the first question, there are n1 answers, c1 of them are correct. Therefore, you have n1-c1 ways to fail the test, and c1 ways to proceed.

The second question has n2 answers, c2 of them are correct. For each way to reach this question, you have nn-cn ways to fail the test, and c2 ways to proceed.

Does that help?
You can get a general formula using this approach, too.
 
  • #3


Thanks for the reply. After I posted this I actually worked out a formula, but it's really clunky and I couldn't really say how I derived it. Here's what I have (by the way I'm uploading a JPG for this because I know if I tried to type it in I would screw it up real nice).

Explanation of notation:
M = total number of answer possibilities (number of "paths")
k = 1,...,K is the number of items on the test
V is the total number of answer choices on item k
VkP is the number of passing answers on item k
VkF is the number of failing answers on item k
Hence, VK is the total number of answer choices on the last (Kth) item on the test, and V1F is the number of failing answers on the first item (k = 1) on the test. (And in the last term in the formula, n is used as an arbitrary index.)

Any advice for making this prettier/more efficient?

Thanks!


You can calculate it step by step: For the first question, there are n1 answers, c1 of them are correct. Therefore, you have n1-c1 ways to fail the test, and c1 ways to proceed.

The second question has n2 answers, c2 of them are correct. For each way to reach this question, you have nn-cn ways to fail the test, and c2 ways to proceed.

Does that help?
You can get a general formula using this approach, too.
 

Attachments

  • #4
34,387
10,475


I did not check all indices, but the formula looks good. I think you can include V_1^F in the last expression, using the convention that an empty product (from n=1 to n=0) is 1.

In the same way, you can include the case "n-1 correct answers and failed at the last one" there, which simplifies the first part a bit.
 
  • #5


Great, thanks for the tips!

I did not check all indices, but the formula looks good. I think you can include V_1^F in the last expression, using the convention that an empty product (from n=1 to n=0) is 1.

In the same way, you can include the case "n-1 correct answers and failed at the last one" there, which simplifies the first part a bit.
 

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