Probability Multiple Choice Item Response

[fleet street]

My school is doing a probability item response SAC on Wednesday. Basically, the idea is that we get 4 multiple choice questions over 50 minutes. For each question, full working must be shown and the errors in reasoning made to get 3 of the 4 incorrect choices must be explained.

Example Question:

Spoiler

If X is a binomial random variable with parameters n=20 and p=0.3, then the mean and variance of X are closest to:
A.
B.
C.
D.
E.

Example Solution:

Spoiler

The answer is A

Incorrect Responses
B: Incorrect assumption that
C: The values of and are switched
D: Incorrect assumption that and The values of and are switched
E: has been calculated instead of

Because of the nature of this SAC, there are few multiple choice questions which provide detailed solutions in this format.
I have a few questions that I would like help trying to discover what the multiple choices are supposed to represent.
Also, I would be happy to try to help anyone else doing a SAC like this if you post your question.

The question that I need help with is attached. Specifically, I need help with A,B and C. (The answer is D and I have already worked out the error in E.)

Thanks for following this post! I know that I haven't written it so that it is easily understandable

[TrueTears]

Since is a pmf and given that the mode is 0 then which implies , but clearly , hence .

[fleet street]

Thanks! But what I'm trying to get at is: "Why would someone put 0<q<0.4?"

[psyxwar]

Thanks! But what I'm trying to get at is: "Why would someone put 0<q<0.4?"

Nossal offers methods in yr10?

[TrueTears]

Thanks! But what I'm trying to get at is: "Why would someone put 0<q<0.4?"

I don't understand. 0<q<0.4 is incorrect.

[fleet street]

I don't understand. 0<q<0.4 is incorrect.

I know. I need to know how someone, using faulty reasoning, would get that answer. Please see the example question that I have given (in spoilers) to see what I'm looking for.

Nossal offers methods in yr10?

Kind of. I've been accepted to do it early, but it isn't usually offered. About 10 other kids do languages 3/4 in year 10.

[TrueTears]

I know. I need to know how someone, using faulty reasoning, would get that answer. Please see the example question that I have given (in spoilers) to see what I'm looking for.

I use correct reasoning.

[fleet street]

I use correct reasoning.

Good point! I'm kind of annoyed at this SAC too.

[BubbleWrapMan]

I know. I need to know how someone, using faulty reasoning, would get that answer.

Although thinking about this is good, you're not always going to be able to determine how the incorrect answers were generated, since they might have just been thought of on the spot without much thought other than "it looks sort of correct".

Perhaps more interesting is to consider that you can eliminate A, B, and C just by looking at all the options. There is only one correct answer, but if A is true then D is also true by transitivity (i.e. if q = 0.4 then 0 < q < 0.8 also holds). Option A being correct means there are two correct answers, which can't happen, so A must be false.

If B is true, then D is also true by similar argument, so B being true leads to two correct answers. Hence B is false.

If C is true, then B and D are also true. Hence C is false.

[Planck's constant]

I need to know how someone, using faulty reasoning, would get that answer.

Get real, my friend.
TrueTears has not used faulty reasoning in maths since he was 3-months old

[fleet street]

Although thinking about this is good, you're not always going to be able to determine how the incorrect answers were generated, since they might have just been thought of on the spot without much thought other than "it looks sort of correct".

Perhaps more interesting is to consider that you can eliminate A, B, and C just by looking at all the options. There is only one correct answer, but if A is true then D is also true by transitivity (i.e. if q = 0.4 then 0 < q < 0.8 also holds). Option A being correct means there are two correct answers, which can't happen, so A must be false.

If B is true, then D is also true by similar argument, so B being true leads to two correct answers. Hence B is false.

If C is true, then B and D are also true. Hence C is false.

Good point, I'll keep this in mind. Also, your argument is interesting.

Get real, my friend.
TrueTears has not used faulty reasoning in maths since he was 3-months old

Speaking of faulty reasoning, any mistake here?