Maths challenge
Author  Message 
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.

Back to Top 
View Profile
Send PM

JQPublic
Posts: 1913

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 
Back to Top 
View Profile
Send PM
Visit Wiki

JQPublic
Posts: 1913

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 
Back to Top 
View Profile
Send PM
Visit Wiki

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.

Back to Top 
View Profile
Send PM

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.

Back to Top 
View Profile
Send PM

JQPublic
Posts: 1913

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 
Back to Top 
View Profile
Send PM
Visit Wiki

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?

Back to Top 
View Profile
Send PM

Skip to Page: 1 ... 4 5 6
 
