The last piece of the puzzle for you guys is combinatorics. Better known as permutations and combinations. There is awesome notes on that
here .
The main thing to remember is, the number of permutation arrangements is order dependent. Combinations are not. I never did find any good tricks/ways to remember that, post any tips below! You also need to know factorials. These are explained in the notes, and are quite simple once you grasp the concept. The formulae are:
where n is the total number of elements, and k is the number of elements we want to arrange.
Applying these formulae is just substitution, so I won't give an example of that. I want to tackle this infamous question from the 2011 extension paper. For those who don't know, 2011's paper was infamously difficult and caused a bit of an uproar. This question was featured by the Sydney Morning Herald. Supposedly extremely difficult. Not for us!
A game is played by throwing darts at a target. A player can choose to throw two or three darts. Darcy plays two games. In Game 1, he chooses to throw two darts, and wins if he hits the target at least once. In Game 2, he chooses to throw three darts, and wins if he hits the target at least twice. The probability that Darcy hits the target on any throw is p, where 0 < p < 1.
(i) Show that the probability that Darcy wins Game 1 is .This first part is simple use of the complimentary result theorem, don't be scared by the variable!
(ii) Show that the probaility that Darcy wins Game 2 is .Similar method as above works a charm, but slightly trickier logic. We have to consider both when he doesn't hit at all, and when he hits once:
The last part of that expression might be a bit confusing. We consider the probability of one hit and two misses, that is:
However, that hit can occur on the first, second, or third throw. This tells us there are three possible arrangements if one hit occurs, all equally likely, so we multiply by three and expand to get the total probability as appears above.
(iii) Prove that Darcy is more likely to win Game 1 than Game 2. This is a bit of an algebra trick more than anything. Now normally this is a big no no, but start with your final result and work backwards:
which is true since 0<p<1, so the original condition holds. Note that this is a perfectly acceptable way to approach proofs in mathematics and extension, and indeed beyond. You just have to be EXTREMELY careful that you don't use any assumptions in your working, and are clear and logical in your argument.
See! Not such a hard question SMH ;) Extension probability can be a doozy, but it is in my opinion, worlds better than projectile motion or binomial proofs. The last question of my HSC paper was probability, and I cheered. It's easy math! Just think carefully, and go slow, and the answers will reveal themselves. Oh! And remember that circular arrangements use (n-1)!, not just n!. Everyone else will forget it, you won't!