BMAT paper 2009 (UK)

[Chwirkytheappleboy]

I did the same method as you, though less random. I started from chart A, bar 1, then worked my way through. :L

Mmmm no you can't do it bar by bar. Because you don't know which two bars are wrong on each chart, it's entirely possible that absolutely everyone got a particular bar wrong except for the teacher (e.g. if E was the correct chart, then everyone else got Bar A wrong). If you're trying to handle it one bar at a time, how can you deal with that situation? You can't really... You have to compare the whole chart.

[a-parsons]

That's a good point. I'm pretty sure I chose D though, but it must have just been blind luck. D: lmao.

That's just a slap in the face... :L

Ya know what this one reminds me of? Fallout 3, hacking computers. It's so obvious. If I'd noticed that in the exam, I would have FLOWN through this question. :| lmao

[Chwirkytheappleboy]

What's Fallout 3? Is that a computer game or something?

Anyway I wouldn't have liked to have taken the BMAT. Individually, the questions aren't too hard (especially when you've got unlimited time as we do here) but being faced with all of these questions and a very short space of time in which to do them would be pretty daunting. UKCAT had the same problem though, but didn't require quite so much science-specific knowledge. I'm glad I only had to do the UKCAT!

[Chwirkytheappleboy]

4 - C
x/x+y+z X y/x+y+z = xy/(x+y+z)^2
Please could someone verify this, and expand upon the explanation, as this is very brief, and I do not understand it enough to explain it myself. Hint hint, Entropy with your pretty diagrams? :L

In total there are x+y+z balls (sum of red, blue and yellow balls)

Note that each time we take a ball, we look at it, then replace it. So the total number of balls never changes. Therefore each time we pick a ball, probability of:

Getting a Red ball: x/(x+y+z)
Getting a Blue ball: y/(x+y+z)
Getting a Yellow ball: z/(x+y+z)

So then, what is the probability of getting a Red ball, then a Blue ball? Well since we must get first a red, AND then a blue second, we multiply the probabilities:

P(1st ball Red, 2nd ball Blue) = x/(x+y+z) x y/(x+y+z) = xy/(x+y+x)^2

[Entropy]

Ah, Chwirky, I get S1Q3 now. Thanks... (even though that was probably the stupidest question ever ugh.) During the time constraint, I read it twice, and just flipped the page. Hopefully I put D for the answer.

[a-parsons]

We have been asked by Cambridge Assessment to remove the paper from the site, and I would ask that anyone who has kept a copy delete it, as it is apparently against the invigilation policy for students to keep a copy of the paper.

On that note, were any of you shown a copy of this invigilation policy before sitting the exam? I was not, so I had just assumed that it followed the same rules and regulations as standard A level exams...

[sara7000]

I don't think it was appropriate to do so in the first place as I am sure everyone or most of the candidates sitting the exam tried to look for past papers and found none, so it must have been unaccepted to distribute such papers. But, I appreciate the effort.

[a-parsons]

Well actually, the woman from Cambridge Assessment emailed me and said that she had made a mistake, that it IS okay to distribute the papers, just not until after November 30th.