Probability of Getting 12 Correct Answers in a Random True False Exam

[mitjak]

Reposting this from the stats forum as I didn't realize it wasn't the correct to post at.

I've been struggling with this problem for two days. It is out of the Concepts in Probability and Stochastic Modelling textbook (1.4-13), and it goes like this:

A true false exam has 15 questions of which 5 are true and 10 are false. A student randomly selects seven questions and answers those false. The remainder of the questions the student answers true. The best possible score the student could obtain on this exam is 12 correct answers. What is the probability of this happening?

The problem appears in one of the introductory chapters, in this case the one on combinations and permutations, but to me it seems like the problem is a bit more complex than that. For instance, when choosing the first 7 questions, the student is guessing, so the odds of him picking the right answer are:
$$({\frac{2}{3}})^7$$.

The sample space doesn't change as he goes about guessing the answers, so neither combinations nor permutations apply here, it seems. But then, when he fills out the remaining 8 true answers, he's no longer randomly picking answers but just marking those remaining 8 answers as false.

I am probably overthinking the problem, but at this point I'm stuck. Would greatly appreciate any help.

 

[╔(σ_σ)╝]

Reposting this from the stats forum as I didn't realize it wasn't the correct to post at.

I've been struggling with this problem for two days. It is out of the Concepts in Probability and Stochastic Modelling textbook (1.4-13), and it goes like this:


The problem appears in one of the introductory chapters, in this case the one on combinations and permutations, but to me it seems like the problem is a bit more complex than that. For instance, when choosing the first 7 questions, the student is guessing, so the odds of him picking the right answer are:
$$({\frac{2}{3}})^7$$.

The sample space doesn't change as he goes about guessing the answers, so neither combinations nor permutations apply here, it seems. But then, when he fills out the remaining 8 true answers, he's no longer randomly picking answers but just marking those remaining 8 answers as false.

I am probably overthinking the problem, but at this point I'm stuck. Would greatly appreciate any help.

Hey,
I too am taking an introductory course in Probability and stochastic processes. Anyway, enough of that.

The way I would approach this problem is first asking myself what is the probability that if the student picks 7 answers they are true ?

$$\frac{2}{3}$$ that is the probability that of choosing a false answer that is correct.

The truth of the matter is that, the probability of choosing a false and correct answer changes .

Consider the following,

$$S =\left\{F,F,F,F,F,F,F,F,F,F,T,T,T,T \right\}$$

$$Event(False) = \left\{F,F,F,F,F,F,F,F,F,F \right\}$$


$$P(False) =\frac{\left|Event(False)\right|}{\left|S \right|}= \frac{2}{3}$$

But after the first correct pick what is the size of the sample space and event space for false answer change, why ?

$$S' = \left\{ F,F,F,F,F,F,F,F,F,T,T,T,T \right\}$$
$$Event(False)' = \left\{F,F,F,F,F,F,F,F,F \right\}$$

$$\left|Event(False)' \right| = 9$$

$$\left|S' \right| = 14$$

$$P(False)= \frac{9}{14}$$

 

[Dick]

You want to count all of the possible ways to choose 7 of the 10 false question to correctly answer 'false' in order to get a score of 12. Divide by all of the possible ways to choose 7 of the questions to answer 'false' regardless of whether they are true or false. Use binomial coefficients to count.

 

[mitjak]

You want to count all of the possible ways to choose 7 of the 10 false question to correctly answer 'false' in order to get a score of 12. Divide by all of the possible ways to choose 7 of the questions to answer 'false' regardless of whether they are true or false. Use binomial coefficients to count.

So, you mean:
$$\frac{ \binom{10}{7} }{ \binom{15}{7} }$$

But I don't fully understand why that works :). It almost seems like we're not using the fact that 10/15 answers are false. Does this work because the answers are only true and false, so we only care about either one, with the other being false?

How would this work if there were 4 possible answers to each question, and the student chose all answers as A (out of {A,B,C,D})?

Thanks!

 

[Dick]

So, you mean:
$$\frac{ \binom{10}{7} }{ \binom{15}{7} }$$

But I don't fully understand why that works :). It almost seems like we're not using the fact that 10/15 answers are false. Does this work because the answers are only true and false, so we only care about either one, with the other being false?

How would this work if there were 4 possible answers to each question, and the student chose all answers as A (out of {A,B,C,D})?

Thanks!

It's simple because you have specified the score MUST be 12. That means ALL of the false answers must be to false questions. That's all you need. Once this is true you know the score from the true answers. Sure you are using the 10 in 10/15. That's where the 10 in the numerator comes from. It is a particularly simple problem.

 

[mitjak]

It is a particularly simple problem.

 

[Dick]

I mean in terms of the number of different cases you have to count. There's only one. If the problem had been what's the probability of getting a score of 8 or more instead of exactly 12, it gets messy.

 

[mitjak]

I mean in terms of the number of different cases you have to count. There's only one. If the problem had been what's the probability of getting a score of 8 or more instead of exactly 12, it gets messy.

And I meant it more as a joke that I felt stupid. Which is not at all abnormal .