# Probability of accessibility

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?

• 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 $frac13$? – Henning Makholm Oct 30 '14 at 16:25
• 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. – Henning Makholm Oct 30 '14 at 16:25
• @HenningMakholm 1/2 then because there's road AC and then AB to BC? – user3434743 Oct 30 '14 at 16:38

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.