Quote Originally Posted by jbwtucker View Post
This is a real world problem we're trying to solve, not a test question, so in a way, your help is all the more appreciated!

A student taking a test is presented with a question with five options. Among the 5 options presented, 2 of the 5 represent correct answers. (So, for example, if A, B, C, D and E are the options, it may be that B and E are correct.)

They may choose up to 2 options; they can opt to select only 1 option, but may select 2. They may not select more than 2. (Essentially, think of 5 check boxes: they may not select more than 2, but they do have the option of selecting only 1.)

We know that the probability of selecting both options correct (i.e., marking two check boxes, and both selections are correct) is represented by the following:

\frac{2}{5}\times\frac{1}{4}=\frac{1}{10}

Here's what we need help with. Please remember that they do have the option of only marking 1 out of the 5 check boxes.

  1. What is the probability that they will get only 1 option correct (that either (a) they mark only 1 check box, and it is a correct one, or (b) they mark 2 check boxes, and 1 is right, but 1 is wrong)?
  2. What is the probability that they will get at least 1 option correct (that either (a) they mark only 1 check box, and it is a correct one, (b) they mark 2 check boxes, and 1 of the 2 is correct, or also (c) they mark 2 check boxes, and both are correct)?


Essentially, we need to know the different probabilities for getting only 1 correct, versus getting at least 1 correct ... and we don't know how to account for their option to only mark 1 checkbox, not two.
Your question has no answer without further assumptions, for instance you seem to assume that the student knows nothing about the question and so will tick boxes at random. Is this realistic?

Also there is no decision problem in this, we need some information to make a informed model of how the student decides to tick zero, one or two boxes (or for that matter why not tick more and what marking rule is then applied).

CB