Prob. of passing mult. SELECTION test, w/ & w/o eliminating

[jbwtucker]

A student is taking a multiple selection test. Each test has 5 options, and 2 of the 5 options are correct answers.

The student may make up to 2 selections. He is not required to make 2; he has the option, on any given question, of making only 1 selection. But he may not make more than 2 selections on each question.

All of the following apply to scenarios in which the student makes two selections:
I know that the probability of selecting both of the correct answers is (2/5)(1/4)=1/10.
I know, by mapping out all possible combinations, that the probability of selecting at least one of the correct answers is 7/10.
I know, by mapping out all possible combinations, that the probability of selecting only one of the correct answers, but NOT both, is 6/10.

The above is the full extent of my knowledge and ability in the area of probability.

The following applies to scenarios in which the student may make 2 selections, but may also choose to make only 1 selection:
Does the fact that the student has the option of only making 1 selection affect the probability that he will select at least one of the correct answers, or is it still 7/10?

We maintain the assumption that the student has the option of making either 1 or 2 selections on each question:

If it's possible, I'd appreciate it if you could show me the equation or function you used to solve this, so that I can attempt to use it with different variables.

Thanks!

 

[tkhunny]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

I removed your duplicate questions on the bottom. Let's work on the other one before we tackle this one. Show your work. You MUST know SOMETHING about it. Let's see it, please.

 

[jbwtucker]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

No, I really don't.

I should probably specify that I am not a student in a probability class, nor have I ever taken a probability class. I know virtually nothing about probability. I am just a person who happens to have a need to use these specific probabilities, in these specific scenarios.

So no, I know nothing of how to do this. Everything I know about this problem, I have already stated in the original post.

Also, not sure why you removed the two questions at the end. They were the most important questions, and the distinctions contained within those two questions were key. These are specifically the things I need to find. What is the probability that the student will get a passing score — with and without the ability to eliminate two wrong answers?

 

[tkhunny]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

1) You already posted those two questions. Duplicate postings are discouraged. Go respond to that post.

2) We have a choice, here. If you REALLY know NOTHING, then this conversation is just as much a waste of your time as it is mine and we should part company. If you really would like to learn and to understand, we must assume some sort of foundation. Show it!

We can rewrite textbooks, but if you have one already, that would not be a fruitful exercise. Please demonstrate that you have attended at least one class or maybe read one chapter. You did get the problem statements from somewhere...

 

[jbwtucker]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

1) That was a completely different context. That was a question in which the student was taking a multiple choice test. Question had 4 options, only 1 of which is a correct answer, and students could only select 1.

THIS one is a multiple selection test. Each questoin has 5 options, 2 of which are correct answers, and the student can select up to 2 of them. Think of it like check boxes instead of radio buttons — if my options are A, B, C, D and E, I'm allowed to make more than one selection. I can select B and E as the two answers I think are correct. However, I'm not required to select 2 answers. I could simply select B as my answer.

The parameters are very different, which is why I asked two separate questions. It's my assumption that they're going to have to be solved differently. Since the contexts of these questions are completely different, it is necessary to ask them again in this context. Essentially, what I'm saying is, "This that and the other factors have changed — now how do I solve it?"

2) I really know nothing. Why is this a waste of time? I need math help, that's why I'm here. Is math help reserved for mathematicians? This kinda of math is at times necessary to use in the real world, but non-mathematicians like myself. When that happens, it'd be really great if those with the skills could help. If you asked a grammar question, or a theology question (two areas in which I'm highly knowledgable), I certainly wouldn't turn you away simply because you haven't demonstrated that you have a level of expertise in the area. I'd be glad to answer your question.

I'd like to understand it. If it's beyond my understanding, I'd at least like to know the steps to be able to perform this specific calculation. I'm better with math than 99% of the general population, I'm just not a mathematician. I'm pretty sure I could follow instructions well enough.

If understanding and knowing the steps to perform it are both beyond me, I'd settle for having a friendly expert on the subject matter tell me what it equals. I understand the reluctance to provide an answer when dealing with students, but I am not a student. I'm just a person with a real world need to know these probabilities, unable to do it myself and hoping someone will be kind enough to help.

Please demonstrate that you have attended at least one class or maybe read one chapter. You did get the problem statements from somewhere...

I have not attended one probability class or read one chapter. I did get the question from somewhere. As I stated above, I got it from real life.

I need to create a test that students will take. This is a diagnostic test, meaning we are solely concerned with discovering what the students DO KNOW. It is important to minimize the extent to which random guessing (i.e., luck/chance) and test taking skills (eliminating wrong answers and then taking a higher % guess) are reflected in the scores. To the greatest extent possible, the scores should reflect only what the test taker does know.

I have a couple options with regards to how to format the test. Option 1 is standard multiple choice questions (4 options, only 1 of which is correct, and only one option can be selected by the student), as stated in my other post. Option 2 is custom multiple selection questions (5 options, 2 of which are correct, with the student being allowed to select either 1 option or 2 different options), as described in this post. I need to know the probability that a student will be successful based purely on (a) chance (guessing), and test taking skills (eliminating and then guessing). That way, I can choose to use the format which provides the LOWEST probability for chance- and test taking skill-based success.

The need to compare the effects of both chance (guessing with no elimination) and test taking skills (guessing with elimination) IN BOTH TEST FORMATS is why I post those same 2 problems (with and without elimination) in both posts (the two different contexts).

This is actually my situation. I "came across" this problem in that it is a need I actually have in real life. It is not a book problem.

Help me, please?

 

[tkhunny]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

I am sorry that you are most likely to respond to what I am about to say as austerity or job security.

You are not prepared for the level of mathematics you need. You need to take a few classes. Algebra I and II and Basic Statistics. Then you'll be ready for it. If you want to jump ahead, look up the "Binomial Thoerem", but you'll need that algebra to deal with it.

Have you ever opened a phrase book in a foreign language? If you find the phrase, "Where's the restroom?" That may be important, but if you cannot understand the reply, what good will it do you? In your application, if someone simply shows you some trick, how will you know if you have applied it correctly? If some student challenges your methodolgy, how shall you defend yourself? There ARE liability issues, here.

#1 If you would spend more time accepting help from the professionals whom you have enlisted, and less time arguing with them, we will all get along a lot better and you may learn something useful.

 

[JeffM]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

The people who are here are volunteers. Most have jobs. All have responsibilities. We cannot give classes and do not do homework. We cannot know what you know and what your situation is unless you tell us.

The formulas for probability theory are few and simple.
0 \(\displaystyle \leq\) P(anything) \(\displaystyle \leq\) 1, where P means the probability of whatever is in parentheses or brackets.
P(an impossibility) = 0.
P(a certainty) = 1.
A and B are mutually exclusive if and only if P(A and B) = 0.
P(A or B) = P(A) + P(B) - P(A and B), where "or" means A or B or both.
The probability of A given that B happens is shown as P(A | B) and is meaningful only if P(B) > 0.
P(A | B) * P(B) = P(A and B) = P(B | A) * P(A).

That's it for basic formulas. The trick comes in manipulating them.

Let's see how they work on the probabilities that you already know in your problem.

First off the implied assumption is that the student knows nothing about the answers and is just making random guesses.
In that case, the probability that the first random guess, guess A, is right is 2/5 because there are 2 correct answers, and a random guess will hit one of those correct answers 2 times out of 5. Make sense? So P(A is right) = (2/5).

What is P(A is wrong)?
Well A is right and A is wrong are mutually exclusive, right? Can't be right AND wrong.
So P(A is right and A is wrong) = 0.
And A is right or A is wrong is certain, right? If it is not right, it is necessarily wrong. If it is not wrong, it is necessarily right. Those are the only choices there are.
So, 1 = P(A is right or A is wrong) = P(A is right) + P(A is wrong) - P(A is right and A is wrong) = (2/5) + P(A is wrong) - 0.
So 1 - (2/5) = 3/5 = P(A is wrong).

What about the second random guess? Well, it depends on the first guess.
If guess A was right, then on guess B there are 3 wrong answers left and 1 right answer. So P(B is right | A is right) = 1/4.
And P(B is wrong | A is right) = 3/4.
If guess A was wrong, then on guess B there are 2 wrong answers left and 2 right answers. P(B is right | A is wrong) = 2/4 = 1/2.
And P(B is wrong | A is wrong) = 2/4 = 1/2.

Now let's use the formulas.
P(A is right and B is right) = P(B is right | A is right) * P(A is right)= (1/4) * (2/5) = 2/20 = 1/10.
P(A is wrong and B is right) = P(B is right | A is wrong) * P(A is wrong) = (1/2) * (3/5) = 3/10.
P(A is right and B is wrong) = P(B is wrong | A is right) * P(A is right) = (3/4) * (2/5) = 6/20 = 3/10.
P(A is wrong and B is wrong) = P(B is wrong | A is wrong) * P(A is wrong) = (1/2) * (3/5) = 3/10.
P(B is right) = P[(A is right and B is right) or (A is wrong and B is right)] =
P(A is right and B is right) + P(A is wrong and B is right) - P(A is right and wrong and B is right) =
(1/10) + (3/10) - 0 = 4/10 = 2/5.
1 = P(B is right or B is wrong) = P(B is right) + P(B is wrong) - P(B is right and wrong) = 2/5 + P(B is wrong) - 0.
So P(B is wrong) = 1 - (2/5) = 3/5.

P(Both are right) = P(A is right and B is right) = 1/10.
P(At least one is right) can be computed three ways using the formulas.
P(At least one is right) = 1 - P(both are wrong) = 1 - (3/10) = 7/10.
P(At least one is right) = P(A is right or B is right) = P(A is right) + P(B is right) - P(A is right and B is right) = (2/5) + (2/5) - (1/10) = 7/10.
P(A is right but not B or B is right but not A or both are right) = (3/10) + (3/10) + (1/10) = 7/10.

It's not the formulas that are hard. It's figuring out which formulas to use when that can be hard.

 

[tkhunny]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

...and I'm going to stick my neck out there and suggest that the OP understood none of that and will find none of it uesful. Oh well, we do what we can.

 

[jbwtucker]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

tkhunny, you're a bit pompouss, aren't you?

As a matter of fact, I understood every single word from JeffM's post, and I thank him profusely for it. The simplicity of it is quite brilliant, actually.

If you would spend more time accepting help from the professionals whom you have enlisted, and less time arguing with them, we will all get along a lot better and you may learn something useful.

I'd love to accept help. If I could get any help at all, I'd not only accept it, I'd express extreme gratitude for it. But until JeffM showed up, I couldn't get any. Three different math forums, and all any of you guys (JeffM excepted, of course) can do is to ask where I got the problem and tell me to take a math class. Nobody will help me.

It's funny, because for just about anything else, you find the people who know and you ask them. You want to know why your computer is behaving the way it is, and how you can fix it? Plenty of people willing to help you out, walk you through the things to check for, and explain how to fix it. No one is going to tell you to go to the local community college and take a class. Need to make a chocolate souffle for your upcoming dinner party? No one's going to tell you to take a cooking class. Need to know if you're using a particular word in the correct context? No one's going to tell you to take an English class. Want to know the details of gun laws in a particular state? No one's going to tell you to go to that state and take a gun safety course. Want to build your own computer, but don't even know what RAM is? There are plenty of people more than willing to help you out. Some will just tell you what they think you should get. Some will explain why you want X amount of RAM, and this processor over that one, without actually teaching you the technical details of how it all works, and some will probably actually get into the gory details and give you a computer hardware lesson on the spot. But no one's going to tell you to go take a computer science class.

No matter what you need to know, if you can't figure it out on your own, given the resources of the internet, then you go find the people that know. There's always someone more than glad to assist you.

Except math, of course. Go find the people who know how to do math and ask them a problem, and it's, Go take a class or four. You want to talk to me about everyone getting along well? Hey, buddy— the rest of us, out here, from every possible walk of life and spanning every possible interest, hobby and discipline? Yeah, we get along swimmingly. Of course, we're not the ones telling anyone who needs help to go take a class. Maybe that has something to do with it...

The really ironic thing is that mathematicians are constantly trying to convince the 98% of the world that hates math and wants to do only as much as is required and not a single iota more that math is useful, practical, necessary, and can and should be frequently applied in real life. But when someone comes in with a real life application that would benefit greatly from the math you guys have been trying to sell everyone since the dawn of time, the people that are in the position to help them make using math a positive experience scorn them, chastise them, and laughed them off.

You want to talk about saving time? You could have saved a lot of time if, like JeffM, you had just helped me — it is what you claim to be here for — instead of wasting so much energy trying to explain why you were too good to bother.

And you're a moderator here? Wow. Can't fathom know who would want to hang around a place where guys like you are setting the tone. Hey dude, a friendly word of advice: Never, ever go into any kind of career that requires even the least bit of customer service — even the least bit of customer interaction!

You're too good for the rest of us.

 

[jbwtucker]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

By the way, tkhunny—

I do have Algebra I and II. I am actually extremely strong in those areas. I could walk in to a college Algebra final exam and get an A. Not that you bothered to ask; just full of assumptions, aren't you?

What I said was that I had very limited knowledge of PROBABILITY. See, here, I'll quote myself:

The above is the full extent of my knowledge and ability in the area of probability (emphasis added).

 

[jbwtucker]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

JeffM,

You, sir, are a gentleman and a scholar! Your explanation was bloody brilliant, and I couldn't possibly express how appreciative I am of you, not only for the help you provided, but also for the willing attitude in which you did so. What a breath of fresh air, after a day spent dealing with, well, pretty much everyone else that hangs out on math help forums.

I wanted to clarify the next step. What you outlined seems to apply to circumstances in which you are always making two selections. How would P(getting at least one correct) change if the student had the option of selecting either one or two answers, and we didn't know which approach he might take? Does it change things at all, or is the probability of getting at least one still going to be 7/10?

Again, thank you so much for your help.

 

[JeffM]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

JeffM,

You, sir, are a gentleman and a scholar! Your explanation was bloody brilliant, and I couldn't possibly express how appreciative I am of you, not only for the help you provided, but also for the willing attitude in which you did so. What a breath of fresh air, after a day spent dealing with, well, pretty much everyone else that hangs out on math help forums.

I wanted to clarify the next step. What you outlined seems to apply to circumstances in which you are always making two selections. How would P(getting at least one correct) change if the student had the option of selecting either one or two answers, and we didn't know which approach he might take? Does it change things at all, or is the probability of getting at least one still going to be 7/10?

Again, thank you so much for your help.

No scholar. But you're welcome.

The short answer to your question is No.

In my previous post, I was trying to give an example of how the basic formulas of probability theory are used in the context of a problem that you already knew. Those basic formulas essentially let you compute compound probabilities IF YOU KNOW the probabilities of the primitive elements. Notice that I assumed that the student was making purely random guesses, which allowed me to say that P(A is right) = 2/5. If the student knew something about the topic, the answers would not be random and P(A is right) > 2/5. In that case, unless you know or have an estimate of what P(A is right) is, the rules for building compound probabilities are useless. (Well there is a theory of subjective probabilites that provides a disciplined basis for thinking about things where you have subjective degrees of belief, but little or no objective data.) So the follow-up question as you posed it has no objective answer unless you can say what is the probability of the student making a single choice and what is the probability of the student making the double choice.

What you can do is to answer the question of whether the student should make the single or the double choice. If I understand the question, the student needs to select a right answer out of five possible answers, only two of which are correct. The student, however, has the right to make one selection or two. Even if the student is making purely random guesses, the student should definitely make the double choice. As we saw before, the probability that a student with no knowledge at all would guess at least one right answer with two choices is 70%. But we also saw that a student with no knowledge at all would have only a 40% chance of randomly guessing the right answer on the first choice (in the example P(A is right) = 2/5, remember). If the student makes only one choice, the first choice is the only choice. So the rational student would always choose twice. Unless the probability that the student will select correctly the first time is 100%, the probability of making at least 1 correct selection increases with two selections rather than one selection. If you understand that, you now have an intuition about why sampling gives increasingly good (but not perfect) information as sample sizes increase.

I hope this helps.

By the way, I lied. Those rules I gave are for basic probability theory. Once you get into more advanced topics like sampling theory and hypothesis testing, you need to learn a bunch more formulas. Most of them are extremely ugly, but you seldom calculate them by hand: you usually use tables or calculators. Moreover, it becomes harder to know which formula is the right one for the job. When I have a problem of that sort, I ask a friend who is a statistician. I don't do brain surgery either although I can put a band aid on. But basic probability theory is simple and elegant.

Now I'd like to put in a word for tkhunny. 80% of the questions we get here are from kids who just want us to do their homework for them in a course that they are taking. Most of us will not do that because just handing out answers means that they do not grasp the concepts and so cannot pass their tests or use the stuff learned in real life. So we tend to be a bit coy, trying to give out hints and suggestions that may lead to understanding. We also get lots of kids who are not prepared for the course that they are taking. Telling them to switch courses so that they don't fail is a kindly thing to do.

 

[tkhunny]

Re: Prob. of passing mult. SELECTION test, w/ & w/o eliminat

You still haven't answered my questions on the other thread. Give it a try. You might learn something. :?

Note: I'm not a moderator because I am cute and cuddly. Generally, it's because I can delete and block pornography, vulgarity, and advertising and keep this a safe place to hang out and study mathematics. That's why I edited your rant for content. You are entitiled to your opinion, but it doesn't seem to be helping even you learn any mathematics. That's why I locked the thread.

I appreciate your examples. Here's one of my own, related to the foreign language idea. You can't give a monkey a computer and ask him to type a legal document for presentation in court. There are reasons why there are prerequisites for mathematics courses. Skipping ahead is very unlikely to be successful.

"pompous" - nice. I don't usually get that one. :wink: