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CaptainBlack · #6 October 26th, 2009, 04:58 AM

Quote:

Originally Posted by instantaneous

wow!

eehhm, so the probability of both happening 3 > a/b > 1 is

p(a>b) * p(3b>a) = 1/2 * 5/6 = 5/12?

So that is one question rather than two?

(Those evenys are not independent so no its not the product).

Assume $a$ is known, then:

$p(b<a \text{ and }b>a/3|a)=2a/3$

(draw a diagram showing the range of $b$ that satisfies the condition)

Then

$p(b<a \text{ and }b>a/3)=\int_{a=0}^1 2a/3\; da$

CB