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was a typo, should be p(X<9) = 1 - p(X>=9) just like I did with the P(X<3) term - hope that helps

P(X<3|X<9) = P( (X<3) and (X<9) ) / P(X<9) = P(X<3)/P(X<9) = (1 - P(X >=3))/(1 - P(X> =9)) = (1-m)/(1-n) = (m-1)/(n-1)

I understand where you're coming from. Due to the time taken to attain this maximum range (with theta = 45) you have to realise that the target is ALSO moving and will have its own relative position given the time of flight for both particles. So the heuristic of plugging in theta = 45 no longer...

Not difficult tbh, SCOM has some harder papers (I used to write papers there)

You are missing a few arrangements, for example the males and females could all be sitting in groups (MMMMFFFF) and still be sitting opposite. The answer would instead be 6*4*2

We have that OP = 2/3 OA and suppose that PQ = k PC. Then OQ = OP + PQ = 2/3 OA + k PC = 2/3 OA + k (OC - OP) = 2/3 OA + k (OC - 2/3 OA). But OQ =lambda OB = lambda (OA + OC). So you can now equate the coefficients of OA and OC in : lambda (OA + OC) = 2/3 OA + k (OC - 2/3 OA). Can supply...

The integral is wrong, you should get -ln|1-y|

Use integration by parts with v' = 1. You will get the following calculation: Hope that helps.

There are (n-1)! ways to arrange n objects on a circle. But arrangement on a ring corresponds to 2 arrangements on the circle (due to the flip). So there are less distinct arrangements on a ring.

Long Division is an absolute trek though - I would strongly advise against it

Gauge how hard the paper is by skim reading every question, then skip straight to any of the "Harder 3U Questions" looking out especially for probability questions, circle geo and inequalities - since often they have a really nice and elegant method of solving. Look at them, and think of how you...