PDF – BAD MATHEMATICS

We’re not actively looking for WitCHes right now, since we have a huge backlog to update. This one, however, came up in another context and, after chatting about it with commenter Red Five, there seemed no choice. The following 1-mark multiple choice question appeared in 2019 Exam 2 (CAS) of VCE’s Mathematical Methods.

The problem was to determine Pr(X > 0), the possible answers being

A. 2/3 B. 3/4 C. 4/5 D. 7/9 E. 5/6

Have fun.

Update (04/07/20)

Who writes this crap? Who writes such a problem, who proofreads such a problem, and then says “Yep, that’ll work”? Because it didn’t work, and it was never going to. The examination report indicates that 27% of students gave the correct answer, a tick or two above random guessing.

We’ll outline a solution below, but first to the crap. The main awfulness is the double-function nonsense, defining the probability distribution $\boldsymbol{f}$

in terms of pretty make the same function $\boldsymbol{p}$

. What’s the point of that? Well, of course $\boldsymbol{f}$

is defined on all of $\boldsymbol{R}$

and $\boldsymbol{p}$

is only defined on $\boldsymbol{[-a,b]}$

. And, what’s the point of defining $\boldsymbol{f}$

on all of $\boldsymbol{R}$

? There’s absolutely none. It’s completely gratuitous and, here, completely ridiculous. It is all the worse, and all the more ridiculous, since the function $\boldsymbol{p}$

isn’t properly defined or labelled piecewise linear, or anything; it’s just Magritte crap.

To add to the Magritte crap, commenter Oliver Oliver has pointed out the hilarious Dali crap, that the Magritte graph is impossible even on its own terms. Beginning in the first quadrant, the point $\boldsymbol{(b,b)}$

is not quite symmetrically placed to make a $45^{\circ}$

angle. And, yeah, the axes can be scaled differently, but why would one do it here? But now for the Dali: consider the second quadrant and ask yourself, how are the axes scaled there? Taking a hit of acid may assist in answering that one.

Now, finally to the problem. As we indicated, the problem itself is fine, it’s just weird and tricky and hellishly long. And worth 1 mark.

As commenters have pointed out, the problem doesn’t have a whole lot to do with probability. That’s just a scenario to give rise to the two equations,

1) $\boldsymbol{a^2 \ +\ \frac{b}{2}\left(2a+b\right) = 1} \qquad \mbox{(triangle + trapezium = 1).}$

and

2) $\boldsymbol{a + b = \frac43} \qquad \mbox{( average = 3/4).}$

The problem is then to evaluate

*) $\boldsymbol{\frac{b}2(2a + b)} \qquad \mbox{(trapezium).}$

or, equivalently,

**) $\boldsymbol{1 - a^2 \qquad} \mbox{(1 - triangle).}$

The problem is tricky, not least because it feels as if there may be an easy way to avoid the full-blown simultaneous equations. This does not appear to be the case, however. Of course, the VCAA just expects the lobotomised students to push the damn buttons which, one must admit, saves the students from being tricked.

Anyway, for the non-lobotomised among us, the simplest approach seems to be that indicated below, by commenter amca01. First multiply equation (1) by 2 and rearrange, to give

3) $\boldsymbol{a^2 + (a + b)^2 = 2}.$

Then, plugging in (2), we have

4) $\boldsymbol{a^2 = \frac29}.$

That then plugs into **), giving the answer 7/9.

Very nice. And a whole 90 seconds to complete, not counting the time lost making sense of all the crap.

Tag: PDF

WitCH 33: Below Average

Update (04/07/20)