WitCH 73: Independent Thinking

Here is an old Methods one for a change, from the 2011 Exam 2. The examination report gives the answer as E and indicates that 15% of students selected this answer.

UPDATE (02/11/21)

As commenter Sai noted, answer E ignores the possibility that Pr(Q) = 0. The use of the word “exactly” in the question is peculiar; the inclusion of the word fells like a conscious act, but the inclusion does nothing other than score an own goal.

16 Replies to “WitCH 73: Independent Thinking”

1. Sai says:

Suppose and . Then, , which means that , and thus . We assume the events are independent, and we write and arrive at the equation . Would the error in the question stem from the use of the word “exactly”, as leads to a whole other case for independence?

1. marty says:

Exactly.

2. John (No) Friend (of VCAA) says:

Partial first crap:
What the apt does “P and Q will be independent events when” mean? (My emphasis)
I’m guessing that VCAA means “P and Q will always be independent events when”

OK, let and .

Using a Karnaugh table, it’s easy to see that P and Q are independent when .

Case 1: and .

Option A is correct only when so reject.

Option B is correct only when so reject.

Option C is always false (it leads to a negative value of x) so reject.

Option D is correct only when so reject.

Option E is correct only when . But is not always true (because of Case 2: below) so reject.

Second crap: There’s no correct answer if we (reasonably) assume VCAA means “P and Q will always be independent events when”.

Case 2: and so either or .
Because of Case 1 there can be no correct option.

So if VCAA wanted Option E to be the correct answer, it should have stated that .

This completes first crap: Incorrectly worded.

It could have been a good question.

All the commercial and free solutions I’ve seen ‘get’ Option E as well.

3. Red Five says:

I’m not going to try to defend this question (mostly because I can’t), but viewing it from the perspective of a student who has to try to answer it in less than 90 seconds…

Pr(P’ and Q) = Pr(P and Q) so, for the sake of time, take a guess than Pr(P) = Pr(P’) (which is not an awful guess since the events are independent) and therefore Pr(P) = 0.5 and then try to make sense of Section B somehow.

I suspect 15% of students ignored the word “exactly”.

1. marty says:

Thanks, RF. I wouldn’t be surprised if 100% of the students ignored the word “exactly”. Still, 15% is pretty low for an MCQ. The question isn’t easy to do quickly, but when the success is worse than throwing darts, one has to wonder.

1. Red Five says:

True. Although, to be fair, conditional probability questions have often had low success rates, with some being as low as 8%.

I know this because these are the questions I use as in class examples (when I’m allowed to teach the subject, so not since 2016…)

1. John Friend says:

2016 … So you were made the scape-goat for VCAA’s shonky 2016 Report. I can see the headlines now:
“Dedicated teacher made the patsy for VCAA’s errors”
“School admits pressure from VCAA to keep competent teacher away from VCE maths – ‘too embarrassing for VCAA’, it said”

More seriously, there’s a good reason why students have a low success rate with conditional probability and a lot of the probability and statistics content in general: It’s taught by hacks who don’t understand it themselves!!

I’ve been told point-blank by a (former) Head of Maths that “Understanding the maths doesn’t mean you’re a good teacher”. My reply was “I agree. But it doesn’t mean you’re a bad teacher, either. But NOT understanding the maths DOES!” (I should have added “You dumb hack” but this was a long time ago when I was more diplomatic).

Unfortunately, in some schools the Head of Maths does not have a strong maths knowledge (sometimes they’re actually out-of-field in the role) and they feel threatened by those who do. They make nasty decisions, and sometimes some teachers get ‘punished’ as a result for their maths being strong.

RF, I’ve been there (not for a long time now but I have a long memory) and it sucks.

2. John (No) Friend (of VCAA) says:

This was probably the genesis for the 2016 Report to say that exact can mean approximate (“Since the given function was approximately a pdf …”). I can hear it now …

Since the given events were exactly independent, option E is the correct answer.

4. Glen says:

This one is terrible. Words have meaning :(.

1. Banacek Spaces says:

Interestingly, if the implication changes to : “If P and Q are non-empty independent events , then Pr(P • Q) = Pr ( P’ • Q) when . . .?”
does lead to Pr(P) =1/2. And rather quickly.
So it is a case of ” if and only if” . One could always try the other way and it may work.
But of course it might not.

5. Alex says:

I think they were aiming for Pr(P and Q) = Pr(P’ and Q) becoming Pr(P)*Pr(Q) = Pr(P’)*Pr(Q) if P and Q are independent. Assuming Pr(Q) is non-zero, divide both sides by it leaving Pr(P) = Pr(P’) so Pr(P) = 0.5

1. marty says:

Yep, that was the aim. Unfortunately, their aim sucks.

6. Terry Mills says:

Perhaps “exactly” is supposed to mean “only”.

1. marty says:

Perhaps. But it doesn’t.

7. Red Five says:

Marty – fells or feels (in your update)?

1. marty says:

Thank, RF: “feels”. But “fells” has a good ring to it. I’ll leave it.

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