1
$\begingroup$

Between A, B, and C, there are the following highways: A – B, A – C, and B – C. During monsoon, when there is heavy rain, each of the road gets blocked independently with probability $p$.

What is then the probability that C will be accessible from A?

I'd say the probability is 1/3, since there's 3 roads that could be blocked off? Am I correct in that assumption?

$\endgroup$
  • 1
    $\begingroup$ Suppose $p$ is $0$ so no road ever gets blocked. Do you still think the probability of being able to get from A to C is only $frac13$? $\endgroup$ – Henning Makholm Oct 30 '14 at 16:25
  • $\begingroup$ Also, I think the intention is that "C is accessible from A" means that either the AC road or both of the AB and BC roads are open. $\endgroup$ – Henning Makholm Oct 30 '14 at 16:25
  • $\begingroup$ @HenningMakholm 1/2 then because there's road AC and then AB to BC? $\endgroup$ – user3434743 Oct 30 '14 at 16:38
0
$\begingroup$

Path A-B-C is free when A-B and B-C is free: $p_{ABC-free} = (1-p)*(1-p)= (1-p)^2$

So it is blocked: $p_{ABC-blocked} = 1-p_{ABC-free} = 1-(1-p)^2= 1-1+2p-p^2 = p(2-p)$

C is not accessible from A if ABC and AC is blocked:

$p_{blocked} = p_{ABC-blocked} * p_{AC-blocked} = p(2-^2) p = p^2(2-p)$

So it is possible to move from A to C with probability

$p_{free} = 1-p_{blocked} = 1- p^2(2-p)$

Check it out:

if p=0: $p_{free} = 1-0 = 1$

if p=1: $p_{free} = 1-1\cdot 1 = 0$

if p =0.5: $p_{free} = 1- 0.5^2(2-0.5) = 1-0.25\cdot 1.5 = 0,625$ You can see. Here with probability of 0.5 you can travel A-C. But with probability 0.5 you can't. There you have a chance of 0.5*0.5 that bot ways are accessable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.