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CaptainBlack · #4 October 26th, 2009, 03:11 AM

Quote:

Originally Posted by instantaneous

thanks for the answer, regarding the question:

you have two variables a and b that are uniformly distributed between 0 and 1

a and b are independent of each other

you chose randomly a and b

what is the probability to chose a and b so that

a > b
3b > a

My suggestion:

a' = a/(a+b)
b' = 1-a'

3 > a'/(1-a') > 1

so that 3/4 > a' > 1/2

giving a probability of 25%... is that correct? is the transformation to a' possible? if so, how would you prove that?

ty

In the first case $p(a>b)=1/2$ simply by symmetry, that is $p(a>b)=p(b>a)$ and $p(a=b)=0$ .

For the second suppose $a$ is chosen first then $p(3b>a|a)=p(b>a/3|a)=1-a/3$ which is:

$p(b>a/3)=\int_{a=0}^{1}p(b>a/3|a)p(a)\;da=\int_{a=0}^{1} (1-a/3)\; da$

CB