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11 years ago
#61
Hey guys, I'm having a brain freeze here.

In STEP 1 Q11 Part 3, in the markscheme they say p < q < r at the start, where p, q and r are the velocities of A_(n-2), A_(n-1) and A_n.

The way it's stated so casually at the start, assumes it's something obvious. I don't get why though, and it's frustrating, as this assumption at the start will prove the contradiction needed.

DFranklin in particular, any help?
0
11 years ago
#62
I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).
0
11 years ago
#63
(Original post by DFranklin)
I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)
Perfect! If all collisions took place, yet p > q, then you'd get another collision as A_(n-2) would catch A_(n-1), which is a contradiction.

(Original post by The God)
It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).
Yes they could keep trotting along at the same speed without colliding, but don't worry, all I need at the end is to justify p <= q + (lambda)r which is a formality now.

Thanks so much mate, 0
11 years ago
#64
(Original post by SimonM)
...
STEP II Q10

(ii)
By subbing into and : , and .

Using , the time taken by B to reach C's initial position following the first A-B collision = In that time, A travels a distance of Therefore, the distance between A and B after B has collided with C = Note that the speed of approach of A w.r.t. B after B has first collided with C is given by: .
Therefore, the time taken for A to reach B after B has collided with C is given by: So total time between the two collisions of A and B is given by as required.
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11 years ago
#65
(Original post by SimonM)
...
STEP I Q12

Second part
Note that:   And   Therefore:    As , our probability tends to 1, as you would expect. Since it's extremely likely that each of the sections of the three roads are blocked, the chances of having more than one open section is minute. This would need to be the case in our scenario where, in the least case (i.e. third road chosen), one section would definitely have to be open.
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11 years ago
#67
STEP I Q14
Whilst this has been discussed earlier I haven't seen anyone post the actual solution so I'll post mine.

i) P(choosing red on rth go)= (usually I would draw a tree diagram but here it just isn't necessary)
To find the maximum we should differentiate and equate the derivative to zero.
let p=P(choosing red on rth go): take logs of both sides:  Now differentiate both sides:  To find stationary points, let  . Since f(n) can not be 0: Rearranging we see that: and so: ii) A trick in these type of questions where the denominator and the numerator of the probabilities go down by one every time is to notice that the k+1th denominator is canceled by the kth numerator. Therefore all we care about is the first denominator and the last numerator as they are the only ones which don't cancel.
P(choosing red on rth go)= so n should be the smallest it can be.
n+1 is greater than or equal to r so the smallest value of n is:
n=r-1
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11 years ago
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