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. Suppose \mathbb{P}(P) = p and \mathbb{P}(Q) = q. Then, \mathbb{P}(P \cap Q) = \mathbb{P}(P' \cap Q), which means that \mathbb{P}(P \cap Q) + \mathbb{P}(P' \cap Q) = q, and thus 2 \mathbb{P}(P \cap Q) = q. We assume the events are independent, and we write 2pq = q and arrive at the equation q(2p-1)=0. Would the error in the question stem from the use of the word “exactly”, as q=0 leads to a whole other case for independence?

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

    OK, let \displaystyle \Pr(P \cap Q) = x and \displaystyle \Pr(P) = a.

    Using a Karnaugh table, it’s easy to see that P and Q are independent when \displaystyle x = 2xa.

    Case 1: \displaystyle a = \frac{1}{2} and \displaystyle 0 \leq x \leq \frac{1}{2}.

    Option A is correct only when \displaystyle x = \frac{1}{4} so reject.

    Option B is correct only when \displaystyle x = \frac{1}{4} so reject.

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

    Option D is correct only when \displaystyle x = \frac{1}{4} so reject.

    Option E is correct only when \displaystyle a = \frac{1}{2}. But \displaystyle a = \frac{1}{2} is not always true (because of Case 2: \displaystyle x = 0 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: \displaystyle x = 0 and so either \displaystyle a = 0 or \displaystyle a = 1.
    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 \displaystyle \Pr(P) \neq 0.

    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. 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. 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. 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. 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 \displaystyle political 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. 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 \displaystyle approximately exactly independent, option E is the correct answer.

    1. 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.

  4. 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

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