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I was trying to solve this problem. There are three roads that connected town A to town B. road 1 and road 2 have 2 sections, and road 3 has 1 section. The probability that a section will be blocked will bee equal to p, and 0

The road would be blocked if either one of the section is blocked or two sections are blocked.

The blocking of road would happen independent of each other, and the probability of choosing road is uniformly distributed.

I tried to solve this problem by listing the case: P( road 1 is chosen and it is not blocked, and road 2 and 3 are blocked 丨road 1 is chosen and it is not blocked )
P(Road 2 chosen and it is not blocked, and road 1 and 3 are blocked 丨 road 2 is chosen and it is not blocked)
P(Road 3 chosen, and it is not blocked and road 1 and 2 are blocked 丨 road 3 chosen and it is not blocked)
Then, adding the probability up. However, the answer I got is not correct.

The answer indicated that I should give the probability like:
P(random road chosen and chosen road open and unchosen road blocked 丨 chosen road open)

I want to know why it is different from the actual answer

Thank you very much for your reply

Update: I really don't know whether my title of question links correctly to the content, I am very sorry for confusion. Please point out any error

Henry Cai
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  • This is not clear. If you are asking "Given that you choose a road uniformly at random, what is the probability that you select an unblocked one?" then, assuming that sections are blocked (or not ) independent of each other, the answer is $\frac 13\times ((1-p^2)+(1-p^2)+ (1-p))$. Or did you want something else? In any case, the formula you provide in the header does not appear to be true. – lulu Jan 10 '20 at 20:44
  • Thank you, the question said hat the sections blocked are independent of each other, and the road would be chosen uniformally – Henry Cai Jan 10 '20 at 20:45
  • I don't see where in the question it says that the sections are independent of each other. – lulu Jan 10 '20 at 20:46
  • The actual answer is $\frac{p^2(2-p)(4-3p)}{3-2p}$ in simplified form. – Henry Cai Jan 10 '20 at 20:46
  • Sorry, I would add this to the question – Henry Cai Jan 10 '20 at 20:46
  • But, really your question is not clear. I am assuming that a "section" is like a lane, so I need both of them to be blocked to make the road useless. But maybe you meant something else. If a "section" means a portion of the road, then maybe you need both sections to be unblocked. But why guess? Please edit your post for clarity. – lulu Jan 10 '20 at 20:47
  • Sorry for asking unclear question, the road would be blocked if either one of the section is blocked or two sections are blocked. I would amend the question now. Thank you for pointing out the error – Henry Cai Jan 10 '20 at 20:48
  • More to the point, none of my interpretations are compatible with the supposed official answer. Note that if $p=0$ then that formula gives $0$...so if there is no chance that any section is blocked, the formula says it is impossible that you will choose an unblocked road. And if $p=1$ your formula gives $\frac 13$...so if it is certain that every section will be blocked, you still have a $\frac 13$ chance of choosing an unblocked road. I think there is some confusion. – lulu Jan 10 '20 at 20:54
  • @lulu Sorry for the confusion. This question is taken from an exam paper. And it does not explicitly say that whether p can be 0 or 1. However, when p =1, the formula would give a probability equal to 1? I think I will add a restriction that 0<p<1 – Henry Cai Jan 10 '20 at 21:08
  • If $p=1$ your formula gives $\frac {1\times 1 \times 1}{3-2}=1$, agreed. So if it is certain that each sector is blocked, it says that it is certain that you will find an unblocked road. But, perhaps it is computing the probability that you find a blocked road? But that doesn't seem right either. As I say, there is some confusion. – lulu Jan 10 '20 at 21:14
  • I see. due to the contradiction, I will restrict the values of p. Thank you very much for pointing out. – Henry Cai Jan 10 '20 at 21:16

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