Maths challenge
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dewtell
Posts: 33

Posted: 08:22PM Oct 4, 2015 

JQP: The participants definitely do not target each other randomly (that was part 1 of the question). It's been long enough since I asked the question that I no longer have my notes on the probability, but you should assume that C shoots in the air/ground on the first round (because killing either of the other two would put him at a major disadvantage by giving a better marksman the first shot at C), B shoots at A (the bigger threat), and if B misses, A will shoot and kill B (again, the bigger threat). Take the probability calculation from there.

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JQPublic
Posts: 1911

Posted: 08:24AM Oct 6, 2015 

That was... much easier.
P(C survivesB's first shot succeeds) = (1/3) / (1  2/9) = 3/7
P(C survivesB's first shot fails) = 1/3
By theorem of total probability,
P(C survives) = 2/3 * 3/7 + 1/3 * 1/3 = 25/63
'An idea, like a ghost, must be spoken to a little before it will explain itself.'  Charles Dickens 
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JQPublic
Posts: 1911

Posted: 04:48AM Jan 11, 2016 

Wow, I log in a few months later and the last post is still me...
Let's revive this with an easy one.
Prove (A^1)^T = (A^T)^1 where A is an n x n matrix.
'An idea, like a ghost, must be spoken to a little before it will explain itself.'  Charles Dickens 
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dewtell
Posts: 33

Posted: 12:13AM Jan 12, 2016 

Where T is a positive integer? Ok, proof by induction. First of all, multiply both sides by A^T on the left, to form the equivalent identity (A^T)(A^1)^T = I. If T=1, this is trivially true, giving us the base case. Now assume that it is true for some T=k, (A^k)(A^1)^k = I, rewrite the left side as (A^k)*I*(A^1)^k = (A^k)(A*A^1)(A^1)^k = (A^k*A)(A^1*(A^1)^k) = (A^(k+1))(A^1)^(k+1) = I. Then by induction, it is true for all positive integers T.

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dewtell
Posts: 33

Posted: 12:28AM Jan 12, 2016 

Or was ^T supposed to be the transpose operator? In that case, I'm pretty sure we can make use of the identity (A*B)^T = (B^T)*(A^T). Do the same multiplication by A^T on the left, so that the identity to be proven is (A^T)(A^1)^T = I. Applying the identity, the left hand side is equivalent to ((A^1)*A)^T = I^T = I.

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JQPublic
Posts: 1911

Posted: 07:28AM Jan 12, 2016 

dewtell wrote: Or was ^T supposed to be the transpose operator? In that case, I'm pretty sure we can make use of the identity (A*B)^T = (B^T)*(A^T). Do the same multiplication by A^T on the left, so that the identity to be proven is (A^T)(A^1)^T = I. Applying the identity, the left hand side is equivalent to ((A^1)*A)^T = I^T = I.
Oh yes, that was my intention. Yep, that's right, now it's your turn
'An idea, like a ghost, must be spoken to a little before it will explain itself.'  Charles Dickens 
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dewtell
Posts: 33

Posted: 12:26AM Jan 19, 2016 

Ok, lets try this one: White and Black Go stones are identical except for the color. I have three identical cloth bags, the first of which contains two white stones, the second of which contains one of each, and the third of which contains two black stones. I pick a bag at random, add a white stone to it, shuffle the stones blindly, and draw one out  which turns out to be white. If I now draw a second stone at random from the same bag (without replacing the first stone), what is the probability that the second stone is also white?

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