HSC 2015 MX2 Permutations & Combinations Marathon (archive) (2 Viewers)
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Re: 2015 permutation X2 marathon
To go broke starting from $1, he must either:
(1) lose $1 with probability (1-p);
OR
(2) gain $1 with probability p AND then go broke from a new starting point of $2.
That is:
P1 = (1-p) + p.P2
To go broke from a starting point of $2, regardless of how much money he accumulates from there, he must first pass through $1.
The probability of going from $2 to $1 must be the same as going from $1 to nothing, ie. P1
So P2 = P1.P1 (two steps of -$1 are needed)
So:
P1 = (1-p) + p.P12
p.P12 - P1 + (1-p) = 0
Solving the quadratic gives:
P1 = 1
OR
P1 = (1-p)/p
The second equation doesn't give valid probabilities for p ≤ 1/2.
So the value of P1 for p ≤ 1/2 must be 1.
It should be clear that if p = 1, then P1 = 0. The second solution gives this value.
The second solution also gives P1 = 1 when p = 1/2.
So the only way of getting a continuity of P1 values as p changes is if:
P1 = 1 for p ≤ 1/2
P1 = (1-p)/p for p > 1/2
Let Pn be the probability that the man goes broke, given that he currently has $n.
To go broke starting from $1, he must either:
(1) lose $1 with probability (1-p);
OR
(2) gain $1 with probability p AND then go broke from a new starting point of $2.
That is:
P1 = (1-p) + p.P2
To go broke from a starting point of $2, regardless of how much money he accumulates from there, he must first pass through $1.
The probability of going from $2 to $1 must be the same as going from $1 to nothing, ie. P1
So P2 = P1.P1 (two steps of -$1 are needed)
So:
P1 = (1-p) + p.P12
p.P12 - P1 + (1-p) = 0
Solving the quadratic gives:
P1 = 1
OR
P1 = (1-p)/p
The second equation doesn't give valid probabilities for p ≤ 1/2.
So the value of P1 for p ≤ 1/2 must be 1.
It should be clear that if p = 1, then P1 = 0. The second solution gives this value.
The second solution also gives P1 = 1 when p = 1/2.
So the only way of getting a continuity of P1 values as p changes is if:
P1 = 1 for p ≤ 1/2
P1 = (1-p)/p for p > 1/2
Re: 2015 permutation X2 marathon
How can we actually justify / be sure though that P(p) needs to be continuous? Is it just something 'obvious'?
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Re: 2015 permutation X2 marathon
That was my worry too. I know the answer is correct, but I'm not sure how to justify that, other than it seems to be common sense (and yes, I know a lot of things that seem to be common sense turn out not to be true).
Re: 2015 permutation X2 marathon
p and q are real numbers.
p is randomly chosen on the interval -a ≤ p ≤ a
q is randomly chosen on the interval 0 ≤ q ≤ b
What is the probability that the equation x² + px + q = 0 has real roots ?
[Note: This is technically a 2 unit question]
p and q are real numbers.
p is randomly chosen on the interval -a ≤ p ≤ a
q is randomly chosen on the interval 0 ≤ q ≤ b
What is the probability that the equation x² + px + q = 0 has real roots ?
[Note: This is technically a 2 unit question]
Last edited:
awesome-0_4000
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Re: 2015 permutation X2 marathon
In how many ways can a coin be tossed n times such that no two consecutive tosses are heads?
In how many ways can a coin be tossed n times such that no two consecutive tosses are heads?
kawaiipotato
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Re: 2015 permutation X2 marathon
The amount of heads (H) and the amount of tails (T) must fall under the condition H =< T for no consecutive heads to occur.
Consider the case when n T's are thrown (n = even)
= 1 way
n-1 T's and 1 H is thrown
= n!/(n-1)! = n ways
n-2 T's and 2 H is thrown
= (n-1)C(2)
n-3 T's and 3H is thrown
= (n-2)C(3)
.
.
.
n/2 H's and n/2 T's
= (n/2 + 1)C(n/2)
Summing, the total ways = 1 + n + (n-2)C(3) + (n-3)C4 + ... + (n/2 + 1)C(n/2)
= summation from k=0 to k=n/2 of (n+1 -k)C(k)
... which I don't know how to evaluate lol
not even sure if this is right can someone check?
The amount of heads (H) and the amount of tails (T) must fall under the condition H =< T for no consecutive heads to occur.
Consider the case when n T's are thrown (n = even)
= 1 way
n-1 T's and 1 H is thrown
= n!/(n-1)! = n ways
n-2 T's and 2 H is thrown
= (n-1)C(2)
n-3 T's and 3H is thrown
= (n-2)C(3)
.
.
.
n/2 H's and n/2 T's
= (n/2 + 1)C(n/2)
Summing, the total ways = 1 + n + (n-2)C(3) + (n-3)C4 + ... + (n/2 + 1)C(n/2)
= summation from k=0 to k=n/2 of (n+1 -k)C(k)
... which I don't know how to evaluate lol
not even sure if this is right can someone check?
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Re: 2015 permutation X2 marathon
Fewer heads than tails does not guarantee no consecutive heads, e.g. we could have H,H,T,T,T,T,T.The amount of heads (H) and the amount of tails (T) must fall under the condition H =< T for no consecutive heads to occur.
Consider the case when n T's are thrown (n = even)
= 1 way
n-1 T's and 1 H is thrown
= n!/(n-1)! = n ways
n-2 T's and 2 H is thrown
= (n-1)C(2)
n-3 T's and 3H is thrown
= (n-2)C(3)
.
.
.
n/2 H's and n/2 T's
= (n/2 + 1)C(n/2)
Summing, the total ways = 1 + n + (n-2)C(3) + (n-3)C4 + ... + (n/2 + 1)C(n/2)
= summation from k=0 to k=n/2 of (n+1 -k)C(k)
... which I don't know how to evaluate lol
not even sure if this is right can someone check?
kawaiipotato
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Re: 2015 permutation X2 marathon
which would mean it's 6C2 for that example
For those cases I used the insertion method
which would mean it's 6C2 for that example
Re: 2015 permutation X2 marathon
Oh right, I think I misunderstood your method before.
kawaiipotato
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awesome-0_4000
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Re: 2015 permutation X2 marathon
This is the approach I was looking for, although I wasn't sure if suggesting recursion made it too easy..$)
kawaiipotato
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Re: 2015 permutation X2 marathon
Is there a way to simplify the summation? Like starting off with a binomial expansion and then substituting some value in
Re: 2015 permutation X2 marathon
Now my question again:
p and q are real numbers.
p is randomly chosen on the interval -a ≤ p ≤ a
q is randomly chosen on the interval 0 ≤ q ≤ b
What is the probability that the equation x² + px + q = 0 has real roots ?
At its core, this is not hard. It involves discriminant and area under a curve.
There is just one hitch.
Now my question again:
p and q are real numbers.
p is randomly chosen on the interval -a ≤ p ≤ a
q is randomly chosen on the interval 0 ≤ q ≤ b
What is the probability that the equation x² + px + q = 0 has real roots ?
At its core, this is not hard. It involves discriminant and area under a curve.
There is just one hitch.
rand_althor
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Re: 2015 permutation X2 marathon
Question I saw: In a chess tournament, n women and 2n men participated, and each one of them played only one game with everybody else. The ratio of the number of games won by women to the number of games won by men is 7:5. Find the number of men that participated in the tournament if no game was over in a tie.
Answer: 6 men
Question I saw: In a chess tournament, n women and 2n men participated, and each one of them played only one game with everybody else. The ratio of the number of games won by women to the number of games won by men is 7:5. Find the number of men that participated in the tournament if no game was over in a tie.
Answer: 6 men
Drsoccerball
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Re: 2015 permutation X2 marathon
?
Are we assuming theres n is even otherwise one person doesnt play another
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