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indeed so :D

ND, do you have an elegant solution to your problem by any chance? me can't find it :S EDIT: Turtle just told me where that one came from.. no wonder you're grinning " :D "

Many of the published solutions are longer than what it could have been

arnold is harder

But it really depends on the test.

80s 70s is on the border

hmm do some by yourself over the summer holidays and keep yourself ahead of the class assuming you're in year 11

let me explain the other solution in a bit more detail: The chance of Dot winning is constant in each state (1 and 2), it doesn't depend on the number of throws prior to it. Examining state 1, 1/6 of the tosses will sum to 7, which changes the state to (2) 1/36 of the tosses will sum...

try this: a, b> 0 a+b <1 prove (a+1)(b+1)< 1/[1-(a+b)]

PCK + PAK = 90 + 90 = 180

Is this the 'finite state machine' method? define 2 'states' (1) The initial state or when the last toss was not a 7 (2) Last toss was a 7. Let A and B be the probability of Dot winning at state 1 and 2 respectively. A = (29/36)A + (1/6)B ....................(1) B = 1/6 +...

the average of overall school marks is around 40-50% for my school...

I was thinking: consider only the first n coins for each person. it's either BOB has more heads or PENNY has more or EQUAL case 1: the last coin bob tosses does not affect the outcome, bob will have more heads case 2: the last coin of bob does not change the outcome, bob will not...

it's a real question, and.. not that hard(long) if you think about it

(i)50% (ii)50% was almost tempted into summing up binomials..

bah.. must be careful in the HSC then

hmm OLDMAN mentioned three ways.. so there must be some 4U way to do this :P Archman, let's expand that to "Second order linear homogenous non-trivial reccurence relation with constant coefficients"

Would have to dispute that result I got 5/9 for Bob and 4/9 for Penny, and seems to reflect the results from a simulation better than 11/18 and 7/18 the results for bob from the simulation were (100000 trials) 55.518% 55.516% 55.579% 55.674% did it this way: ok.. let the probability...

I think it depends on your current marks Level 1: Couchman Level 2: Fitzy/ Cambridge Level 3: Coroneous 0-69 Level 1 70-89 level 2 90-100 level 3

actually.. you should practice without the aid..