PAT and Other Things
So Wednesday 4th was the Oxford PAT test. This year they removed short and long questions, leaving only the 'medium' ones in the physics paper. Personally, I thought that it was quite a bit easier this year than last year - it's hard to think of a specific question that was particularly interesting.
I do quite like a probability question that came up: toss a fair coin three times. Given that one of the tosses landed tails, what chance is there that all three came up tails? Personally, I think that this is ambiguous, because there are a couple of different meanings:
What Oxford probably meant:
"Hello, I've flipped a coin three times and AT LEAST ONE coin was tails, what chance is there that all of them were tails" Answer: 7/8, because P(3T | 1T) = P(1T | 3T) * P(3T) / P( 1T) by Bayes' rule which is equal to (1)*(1/8)/(7/8)=(7/8).
What Oxford probably did not mean:
"Hello, I've flipped a coin three time. The SECOND TIME it landed tails, what chance is there that all of them were tails?" Answer: 1/4 because P(TTT | ?T?) = P (?T? | TTT) * P (TTT) / P(?T?) again by Bayes' and this equals (1)*(1/8)/(1/2)=(1/4).
What Oxford almost certainly did not mean:
"Hello, I've flipped a coin three times. EXACTLY ONE COIN was tails, what is the probability of all of them being tails?" Answer: 0, because... seriously? Oh, okay: P(3T | 2H) = P(2H |3T) P(3T) / P(2H) is given to us by Bayes, then P(3T |2H) = 0*(1/8)/(3/8)=0.
So the wording does affect the answer and Oxford's wording could conceivably mean any of the three (specifically, if I am supposed to tell you that "one of the flips was tails" does that include telling you which flip it was? Because that's extra information that narrows down the search space.)
On Thursday I did the Senior Maths Challenge and went to Westminster to talk about our science project Helios at the Big Bang fair. This time round, I wasn't so hot on some of the geometry questions, but I still think I did pretty well overall. The last question this year was pretty easy:
f(xy) = f(x) + f(y) ; f(10)=14 , f(40)= 21 (I can't remember the exact numbers, bear with me)
Find f(500)
Firstly, as a check, we know that logs act in this way, and that log(a)<log(b) for a<b. Let's see if this function does as well:
f(10) < f(40) good. f(4) = f(40/10) = f(40)-f(10) = 7 < f(10). So that's good. We can probably guess that this property will be true for all a and b.
Next check: f(400) < f(500) < f(1000) , 35 < f(500) < 42. So we want our answer to be in this range. Actually, on the paper there was only one multiple choice option within that range, so you could be done now.
Let's do this properly:
500 = 10^a * 40^b
f(500)=af(10) + bf(40)
So we just need to find a and b. This can be done with the magic of prime factors and simultaneous equations, which I won't go into here too much especially since I think my original numbers were wrong.
(2^2 * 5^3) = (2*5)^a * (2^3 * 5)^b
This gives us a square system for a and b which can be solved. I won't torture you with that just now though. We also see another nice fact: that so long as we are given f(a) and f(b) with a certain niceness to their prime factorisation, we can find f(c). The niceness? We need a and b to contain at least one factor of every prime in c, and we would like it that they contain no other prime factors. It is obviously possible for them to both have a prime factor that is in c such as f(7) and f(14) being used to make f(2), but this is a much more restrictive situation. In fact, you will always be able to find numbers a and b, but the above restrictions should ensure (I think) that you always get rational answers not involving logs. A lot of the time they should be integers if f(a) and f(b) are.
I have an interview at Exeter in just over a week, and then I'm going to Nottingham for a day at the start of December. But before that... the Physics Olympiad is coming up very soon!
